CBSE Maths › Class 10 PYQs › Introduction to Trigonometry

Introduction to Trigonometry: 3 marks Questions (CBSE Class 10)

86 different 3 marks questions on Introduction to Trigonometry from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (139)2 marks (88)3 marks (86)4 marks (1)

Q283 marksShort AnswerIntroduction to Trigonometry CBSE 2026 · Basic 430/4/1

Prove that $\frac{c o t A - c o s A}{c o t A + c o s A} = \frac{s e c A - t a n A}{s e c A + t a n A}$

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Answer: Proved.

LHS: $\frac{\frac{c o s A}{s i n A} - c o s A}{\frac{c o s A}{s i n A} + c o s A} = \frac{c o s A ( \frac{1 - s i n A}{s i n A} )}{c o s A ( \frac{1 + s i n A}{s i n A} )} = \frac{1 - s i n A}{1 + s i n A}$ .
RHS: $\frac{\frac{1}{c o s A} - \frac{s i n A}{c o s A}}{\frac{1}{c o s A} + \frac{s i n A}{c o s A}} = \frac{1 - s i n A}{1 + s i n A}$ .
LHS = RHS. Hence proved.

Also asked in: 2026 Basic 430/4/3

Q293 marksShort AnswerIntroduction to Trigonometry CBSE 2026 · Basic 430/4/2

Prove that $\frac{( s e c θ + t a n θ ) ^{2} - 1}{( s e c θ + t a n θ ) ^{2} + 1} = sin θ$ .

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Answer: Proved.

Numerator = $sec^{2} θ + 2 sec θ tan θ + tan^{2} θ - 1 = 2 tan^{2} θ + 2 sec θ tan θ = 2 tan θ (tan θ + sec θ)$ (using $sec^{2} θ - 1 = tan^{2} θ$ ).
Denominator = $sec^{2} θ + 2 sec θ tan θ + tan^{2} θ + 1 = 2 sec^{2} θ + 2 sec θ tan θ = 2 sec θ (sec θ + tan θ)$ (using $tan^{2} θ + 1 = sec^{2} θ$ ).
LHS = $\frac{t a n θ}{s e c θ} = \frac{s i n θ}{c o s θ} \times cos θ = sin θ$ = RHS.

Q29 (OR) (OR)3 marksShort AnswerIntroduction to Trigonometry CBSE 2026 · Basic 430/4/2

Prove that $\frac{t a n A}{1 - c o t A} + \frac{c o t A}{1 - t a n A} = 1 + sec A cosec A$ .

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Answer: Proved.

Let $t = tan A$ , so $cot A = \frac{1}{t}$ .
$\frac{t}{1 - \frac{1}{t}} + \frac{\frac{1}{t}}{1 - t} = \frac{t ^{2}}{t - 1} - \frac{1}{t ( t - 1 )} = \frac{t ^{3} - 1}{t ( t - 1 )}$ .
$= \frac{( t - 1 ) ( t ^{2} + t + 1 )}{t ( t - 1 )} = t + 1 + \frac{1}{t} = 1 + tan A + cot A$ .
$tan A + cot A = \frac{s i n ^{2} A + c o s ^{2} A}{s i n A c o s A} = sec A cosec A$ .
Hence LHS = $1 + sec A cosec A$ = RHS.

Q293 marksShort AnswerIntroduction to Trigonometry CBSE 2026 · Basic 430/5/1

Prove that $\frac{1 - s i n A}{1 + s i n A} = \frac{1}{s e c A + t a n A}$ .

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Answer: Proved.

LHS $= \frac{( 1 - s i n A ) ( 1 - s i n A )}{( 1 + s i n A ) ( 1 - s i n A )} = \frac{( 1 - s i n A ) ^{2}}{c o s ^{2} A} = \frac{1 - s i n A}{c o s A}$
$= sec A - tan A$
$= \frac{( s e c A - t a n A ) ( s e c A + t a n A )}{s e c A + t a n A} = \frac{s e c ^{2} A - t a n ^{2} A}{s e c A + t a n A} = \frac{1}{s e c A + t a n A}$ = RHS

Q303 marksShort AnswerIntroduction to Trigonometry CBSE 2026 · Basic 430/5/2

Prove that : $tan^{2} θ + cot^{2} θ + 2 = sec^{2} θ cosec^{2} θ$ .

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Answer: Proved.

LHS $= (1 + tan^{2} θ) + (1 + cot^{2} θ) = sec^{2} θ + cosec^{2} θ$
$= \frac{1}{c o s ^{2} θ} + \frac{1}{s i n ^{2} θ} = \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n ^{2} θ c o s ^{2} θ} = \frac{1}{s i n ^{2} θ c o s ^{2} θ}$
$= sec^{2} θ cosec^{2} θ$ = RHS

Q313 marksShort AnswerIntroduction to Trigonometry CBSE 2026 · Basic 430/5/3

Prove that : $\frac{1 - c o s A}{1 + c o s A} = \frac{t a n A}{s e c A + 1}$ .

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Answer: Proved.

LHS $= \frac{( 1 - c o s A ) ( 1 + c o s A )}{( 1 + c o s A ) ^{2}} = \frac{s i n ^{2} A}{( 1 + c o s A ) ^{2}} = \frac{s i n A}{1 + c o s A}$
RHS $= \frac{\frac{s i n A}{c o s A}}{\frac{1}{c o s A} + 1} = \frac{\frac{s i n A}{c o s A}}{\frac{1 + c o s A}{c o s A}} = \frac{s i n A}{1 + c o s A}$
LHS = RHS. Proved.

Q283 marksShort AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/1/1

If $x = h + a cos θ$ , $y = k + b sin θ$ , then prove that :
$(\frac{x - h}{a})^{2} + (\frac{y - k}{b})^{2} = 1$

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Answer: Proved.

From the given, $\frac{x - h}{a} = cos θ$ and $\frac{y - k}{b} = sin θ$ .
So LHS $= cos^{2} θ + sin^{2} θ = 1 =$ RHS.

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q283 marksShort AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/2/1

Prove that : $\frac{s e c ^{3} θ}{s e c ^{2} θ - 1} + \frac{cose c ^{3} θ}{cose c ^{2} θ - 1} = sec θ . cosec θ (sec θ + cosec θ)$

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Answer: Proved.

$sec^{2} θ - 1 = tan^{2} θ$ and $cose c^{2} θ - 1 = cot^{2} θ$ .
LHS $= \frac{s e c ^{3} θ}{t a n ^{2} θ} + \frac{cose c ^{3} θ}{c o t ^{2} θ} = \frac{1}{c o s ^{3} θ} \cdot \frac{c o s ^{2} θ}{s i n ^{2} θ} + \frac{1}{s i n ^{3} θ} \cdot \frac{s i n ^{2} θ}{c o s ^{2} θ}$
$= \frac{1}{c o s θ s i n ^{2} θ} + \frac{1}{s i n θ c o s ^{2} θ} = \frac{c o s θ + s i n θ}{s i n ^{2} θ c o s ^{2} θ}$
RHS $= \frac{1}{c o s θ s i n θ} (\frac{1}{c o s θ} + \frac{1}{s i n θ}) = \frac{s i n θ + c o s θ}{s i n ^{2} θ c o s ^{2} θ}$
Hence LHS = RHS.

Also asked in: 2026 Standard 30/2/2, 2026 Standard 30/2/3

Q28 (OR) (OR)3 marksShort AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/2/1

If $\frac{s e c α}{cosec β} = p$ and $\frac{t a n α}{cosec β} = q$ , then prove that $(p^{2} - q^{2}) sec^{2} α = p^{2}$ .

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Answer: Proved.

$p = sec α sin β$ and $q = tan α sin β$ .
$p^{2} - q^{2} = sin^{2} β (sec^{2} α - tan^{2} α) = sin^{2} β$
$(p^{2} - q^{2}) sec^{2} α = sin^{2} β sec^{2} α = (sec α sin β)^{2} = p^{2}$
Hence proved.

Also asked in: 2026 Standard 30/2/2, 2026 Standard 30/2/3

Q263 marksShort AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/3/1

If $sin θ + cos θ = 3$ , then prove that
$tan θ + cot θ = 1$

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Answer: Proved.

Square both sides: $sin^{2} θ + cos^{2} θ + 2 sin θ cos θ = 3$
$1 + 2 sin θ cos θ = 3$ , so $sin θ cos θ = 1$
$tan θ + cot θ = \frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ} = \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n θ c o s θ} = \frac{1}{1} = 1$
Hence proved

Also asked in: 2026 Standard 30/3/2, 2026 Standard 30/3/3

Q26 (OR) (OR)3 marksShort AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/3/1

Prove that :
$(sin A + sec A)^{2} + (cos A + cosec A)^{2} = (1 + sec A cosec A)^{2}$

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Answer: Proved.

LHS $= sin^{2} A + 2 sin A sec A + sec^{2} A + cos^{2} A + 2 cos A cosec A + cosec^{2} A$
$= 1 + 2 (tan A + cot A) + sec^{2} A + cosec^{2} A$
$tan A + cot A = \frac{s i n ^{2} A + c o s ^{2} A}{s i n A c o s A} = sec A cosec A$
$sec^{2} A + cosec^{2} A = \frac{s i n ^{2} A + c o s ^{2} A}{s i n ^{2} A c o s ^{2} A} = sec^{2} A cosec^{2} A$
LHS $= 1 + 2 sec A cosec A + sec^{2} A cosec^{2} A = (1 + sec A cosec A)^{2} =$ RHS

Also asked in: 2026 Standard 30/3/2, 2026 Standard 30/3/3

Q303 marksShort AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/4/2

Prove that : $\frac{1}{s e c x - t a n x} - \frac{1}{c o s x} = \frac{1}{c o s x} - \frac{1}{s e c x + t a n x}$

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Answer: Proved.

$\frac{1}{s e c x - t a n x} = \frac{s e c x + t a n x}{s e c ^{2} x - t a n ^{2} x} = sec x + tan x$
LHS $= sec x + tan x - sec x = tan x$
$\frac{1}{s e c x + t a n x} = \frac{s e c x - t a n x}{s e c ^{2} x - t a n ^{2} x} = sec x - tan x$
RHS $= sec x - (sec x - tan x) = tan x$
LHS = RHS

Q313 marksShort AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/4/3

If $sec θ + tan θ = m$ , show that $\frac{m ^{2} - 1}{m ^{2} + 1} = sin θ$ .

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Answer: Proved.

$m = \frac{1 + s i n θ}{c o s θ}$ , so $m^{2} = \frac{( 1 + s i n θ ) ^{2}}{c o s ^{2} θ}$
$\frac{m ^{2} - 1}{m ^{2} + 1} = \frac{( 1 + s i n θ ) ^{2} - c o s ^{2} θ}{( 1 + s i n θ ) ^{2} + c o s ^{2} θ}$
Numerator $= 1 + 2 sin θ + sin^{2} θ - (1 - sin^{2} θ) = 2 sin θ (1 + sin θ)$
Denominator $= 1 + 2 sin θ + sin^{2} θ + cos^{2} θ = 2 (1 + sin θ)$
Ratio $= sin θ$

Q283 marksShort AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/5/1

Prove that : $\frac{t a n θ}{1 - c o t θ} + \frac{c o t θ}{1 - t a n θ} = 1 + tan θ + cot θ$ .

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Answer: Proved.

Let $t = tan θ$ , so $cot θ = \frac{1}{t}$ .
$\frac{t}{1 - \frac{1}{t}} = \frac{t ^{2}}{t - 1}$ and $\frac{1/ t}{1 - t} = - \frac{1}{t ( t - 1 )}$
LHS $= \frac{t ^{2}}{t - 1} - \frac{1}{t ( t - 1 )} = \frac{t ^{3} - 1}{t ( t - 1 )}$
$= \frac{( t - 1 ) ( t ^{2} + t + 1 )}{t ( t - 1 )} = \frac{t ^{2} + t + 1}{t} = t + 1 + \frac{1}{t}$
$= 1 + tan θ + cot θ$ = RHS

Also asked in: 2026 Standard 30/5/3

Q313 marksShort AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/5/2

Prove that : $(1 + cot θ - cosec θ) (1 + tan θ + sec θ) = 2$

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Answer: Proved.

Write in terms of $sin θ$ and $cos θ$ :
$1 + cot θ - cosec θ = \frac{s i n θ + c o s θ - 1}{s i n θ}$
$1 + tan θ + sec θ = \frac{c o s θ + s i n θ + 1}{c o s θ}$
LHS $= \frac{( s i n θ + c o s θ ) ^{2} - 1}{s i n θ c o s θ}$
$= \frac{s i n ^{2} θ + c o s ^{2} θ + 2 s i n θ c o s θ - 1}{s i n θ c o s θ} = \frac{2 s i n θ c o s θ}{s i n θ c o s θ} = 2$ = RHS

Q303 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/1/1

Prove that :
$\frac{1 + c o t ^{2} A}{1 + t a n ^{2} A} = (\frac{1 - c o t A}{1 - t a n A})^{2}$

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Answer: Proved.

LHS $= \frac{cosec ^{2} A}{s e c ^{2} A} = \frac{c o s ^{2} A}{s i n ^{2} A} = cot^{2} A$ .
$\frac{1 - c o t A}{1 - t a n A} = \frac{\frac{s i n A - c o s A}{s i n A}}{\frac{c o s A - s i n A}{c o s A}} = - \frac{c o s A}{s i n A} = - cot A$ .
RHS $= (- cot A)^{2} = cot^{2} A$ = LHS.

Q283 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/1/2

Prove the following trigonometric identity :
$\frac{cosec A - 1}{cosec A + 1} = sec A - tan A$

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Answer: Proved.

$\frac{cosec A - 1}{cosec A + 1} = \frac{\frac{1}{s i n A} - 1}{\frac{1}{s i n A} + 1} = \frac{1 - s i n A}{1 + s i n A}$ .
Multiply numerator and denominator by $(1 - sin A)$ : $\frac{( 1 - s i n A ) ^{2}}{1 - s i n ^{2} A} = \frac{( 1 - s i n A ) ^{2}}{c o s ^{2} A}$ .
Taking the square root (A acute, so $1 - sin A \geq 0$ , $cos A > 0$ ): LHS $= \frac{1 - s i n A}{c o s A}$ .
$= \frac{1}{c o s A} - \frac{s i n A}{c o s A} = sec A - tan A$ = RHS.

Q263 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/1/3

Prove the following trigonometric identity :
$(sin A - cosec A) (cos A - sec A) = \frac{1}{t a n A + c o t A}$

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Answer: Proved.

LHS $= (sin A - \frac{1}{s i n A}) (cos A - \frac{1}{c o s A}) = \frac{s i n ^{2} A - 1}{s i n A} \cdot \frac{c o s ^{2} A - 1}{c o s A}$ .
$= \frac{( - c o s ^{2} A ) ( - s i n ^{2} A )}{s i n A c o s A} = sin A cos A$ .
RHS $= \frac{1}{\frac{s i n A}{c o s A} + \frac{c o s A}{s i n A}} = \frac{s i n A c o s A}{s i n ^{2} A + c o s ^{2} A} = sin A cos A$ .
Hence LHS = RHS.

Q303 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/2/1

Prove the following trigonometric identity :
$\frac{c o s θ}{1 + s i n θ} + \frac{1 + s i n θ}{c o s θ} = 2 sec θ$

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Answer: Proved.

LHS $= \frac{c o s ^{2} θ + ( 1 + s i n θ ) ^{2}}{( 1 + s i n θ ) c o s θ}$ .
Numerator $= cos^{2} θ + 1 + 2 sin θ + sin^{2} θ = 2 + 2 sin θ = 2 (1 + sin θ)$ .
LHS $= \frac{2 ( 1 + s i n θ )}{( 1 + s i n θ ) c o s θ} = \frac{2}{c o s θ} = 2 sec θ$ = RHS.

Q283 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/2/2

Prove the following trigonometric identity :
$\frac{1 - s i n θ}{c o s θ} + \frac{c o s θ}{1 - s i n θ} = 2 sec θ$

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Answer: Proved.

LHS $= \frac{( 1 - s i n θ ) ^{2} + c o s ^{2} θ}{c o s θ ( 1 - s i n θ )}$ .
Numerator $= 1 - 2 sin θ + sin^{2} θ + cos^{2} θ = 2 - 2 sin θ = 2 (1 - sin θ)$ .
LHS $= \frac{2 ( 1 - s i n θ )}{c o s θ ( 1 - s i n θ )} = \frac{2}{c o s θ} = 2 sec θ$ = RHS.

Q263 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/2/3

Prove the following trigonometric identity :
$\frac{c o s A - 2 c o s ^{3} A}{2 s i n ^{3} A - s i n A} = cot A$

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Answer: Proved.

LHS $= \frac{c o s A ( 1 - 2 c o s ^{2} A )}{s i n A ( 2 s i n ^{2} A - 1 )}$ .
$1 - 2 cos^{2} A = 1 - 2 (1 - sin^{2} A) = 2 sin^{2} A - 1$ .
LHS $= \frac{c o s A ( 2 s i n ^{2} A - 1 )}{s i n A ( 2 s i n ^{2} A - 1 )} = \frac{c o s A}{s i n A} = cot A$ = RHS.

Q293 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/3/1

Prove that :
$\frac{1 + cosec A}{cosec A} = \frac{c o s ^{2} A}{1 - s i n A}$

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Answer: Proved.

LHS $= \frac{1 + \frac{1}{s i n A}}{\frac{1}{s i n A}} = sin A + 1$ .
RHS $= \frac{c o s ^{2} A}{1 - s i n A} = \frac{1 - s i n ^{2} A}{1 - s i n A} = \frac{( 1 - s i n A ) ( 1 + s i n A )}{1 - s i n A} = 1 + sin A$ .
LHS = RHS. Hence proved.

Q273 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/3/2

Prove the following trigonometric identity :
$\frac{1 + s e c A}{s e c A} = \frac{s i n ^{2} A}{1 - c o s A}$

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Answer: Proved.

LHS $= \frac{1 + \frac{1}{c o s A}}{\frac{1}{c o s A}} = cos A + 1$ .
RHS $= \frac{1 - c o s ^{2} A}{1 - c o s A} = \frac{( 1 - c o s A ) ( 1 + c o s A )}{1 - c o s A} = 1 + cos A$ .
LHS = RHS. Hence proved.

Q313 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/3/3

Prove the following trigonometric identity :
$\frac{t a n θ}{1 + c o t θ} + \frac{c o t θ}{1 + t a n θ} = tan θ + cot θ - 1$

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Answer: Proved.

Let $t = tan θ$ , so $cot θ = \frac{1}{t}$ .
$\frac{t}{1 + \frac{1}{t}} = \frac{t ^{2}}{t + 1}$ and $\frac{\frac{1}{t}}{1 + t} = \frac{1}{t ( t + 1 )}$ .
LHS $= \frac{t ^{3} + 1}{t ( t + 1 )} = \frac{( t + 1 ) ( t ^{2} - t + 1 )}{t ( t + 1 )} = \frac{t ^{2} - t + 1}{t}$ .
$= t - 1 + \frac{1}{t} = tan θ + cot θ - 1 =$ RHS. Hence proved.

Q283 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/5/1

Prove that $(cosec A + sin A)^{2} + (sec A + cos A)^{2} = 7 + tan^{2} A + cot^{2} A$ .

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Answer: Proved.

LHS $= cosec^{2} A + 2 cosec A sin A + sin^{2} A + sec^{2} A + 2 sec A cos A + cos^{2} A$ .
$cosec A sin A = 1$ and $sec A cos A = 1$ , so LHS $= cosec^{2} A + sec^{2} A + 4 + (sin^{2} A + cos^{2} A)$ .
$= (1 + cot^{2} A) + (1 + tan^{2} A) + 4 + 1$ .
$= 7 + tan^{2} A + cot^{2} A$ = RHS.

Q293 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/5/2

Prove that : $\frac{c o s θ}{1 - t a n θ} + \frac{s i n θ}{1 - c o t θ} = cos θ + sin θ$ .

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Answer: Proved.

$\frac{c o s θ}{1 - t a n θ} = \frac{c o s θ}{1 - \frac{s i n θ}{c o s θ}} = \frac{c o s ^{2} θ}{c o s θ - s i n θ}$ .
$\frac{s i n θ}{1 - c o t θ} = \frac{s i n θ}{1 - \frac{c o s θ}{s i n θ}} = \frac{s i n ^{2} θ}{s i n θ - c o s θ} = - \frac{s i n ^{2} θ}{c o s θ - s i n θ}$ .
LHS $= \frac{c o s ^{2} θ - s i n ^{2} θ}{c o s θ - s i n θ} = \frac{( c o s θ - s i n θ ) ( c o s θ + s i n θ )}{c o s θ - s i n θ}$ .
$= cos θ + sin θ$ = RHS.

Q263 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/5/3

Prove that : $(sin θ + sec θ)^{2} + (cos θ + cosec θ)^{2} = (1 + sec θ cosec θ)^{2}$ .

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Answer: Proved.

LHS $= sin^{2} θ + 2 sin θ sec θ + sec^{2} θ + cos^{2} θ + 2 cos θ cosec θ + cosec^{2} θ$ .
$= 1 + 2 (\frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ}) + sec^{2} θ + cosec^{2} θ$ .
$\frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ} = \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n θ c o s θ} = sec θ cosec θ$ .
$sec^{2} θ + cosec^{2} θ = \frac{1}{c o s ^{2} θ} + \frac{1}{s i n ^{2} θ} = \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n ^{2} θ c o s ^{2} θ} = sec^{2} θ cosec^{2} θ$ .
LHS $= 1 + 2 sec θ cosec θ + sec^{2} θ cosec^{2} θ = (1 + sec θ cosec θ)^{2}$ = RHS.

Q303 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/6/1

Prove that $\frac{t a n θ}{1 - c o t θ} + \frac{c o t θ}{1 - t a n θ} = sec θ cosec θ + 1$ .

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Answer: Proved.

Let $t = tan θ$ , so $cot θ = \frac{1}{t}$ .
LHS $= \frac{t}{1 - \frac{1}{t}} + \frac{\frac{1}{t}}{1 - t} = \frac{t ^{2}}{t - 1} - \frac{1}{t ( t - 1 )} = \frac{t ^{3} - 1}{t ( t - 1 )}$ .
$= \frac{( t - 1 ) ( t ^{2} + t + 1 )}{t ( t - 1 )} = \frac{t ^{2} + t + 1}{t} = t + \frac{1}{t} + 1$ .
$tan θ + cot θ = \frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ} = \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n θ c o s θ} = sec θ cosec θ$ .
So LHS $= sec θ cosec θ + 1$ = RHS.

Q273 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/1/1

Prove that : $\frac{t a n θ}{1 - c o t θ} + \frac{c o t θ}{1 - t a n θ} = 1 + sec θ cosec θ$

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Answer: Proved.

Let $t = tan θ$ , so $cot θ = \frac{1}{t}$ .
LHS $= \frac{t}{1 - \frac{1}{t}} + \frac{\frac{1}{t}}{1 - t} = \frac{t ^{2}}{t - 1} - \frac{1}{t ( t - 1 )}$
$= \frac{t ^{3} - 1}{t ( t - 1 )} = \frac{( t - 1 ) ( t ^{2} + t + 1 )}{t ( t - 1 )} = \frac{t ^{2} + t + 1}{t}$
$= t + 1 + \frac{1}{t} = 1 + tan θ + cot θ$
$tan θ + cot θ = \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n θ c o s θ} = \frac{1}{s i n θ c o s θ} = sec θ cosec θ$ .
So LHS $= 1 + sec θ cosec θ =$ RHS.

Also asked in: 2025 Standard 30/1/2, 2025 Standard 30/1/3

Q273 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/2/1

Prove that : $(1 + \frac{1}{t a n ^{2} θ}) (1 + \frac{1}{c o t ^{2} θ}) = \frac{1}{s i n ^{2} θ - s i n ^{4} θ}$

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Answer: Proved.

LHS $= (1 + cot^{2} θ) (1 + tan^{2} θ)$
$= cosec^{2} θ \cdot sec^{2} θ = \frac{1}{s i n ^{2} θ c o s ^{2} θ}$
$= \frac{1}{s i n ^{2} θ ( 1 - s i n ^{2} θ )} = \frac{1}{s i n ^{2} θ - s i n ^{4} θ}$ = RHS

Also asked in: 2025 Basic 430/6/2

Q27 (OR) (OR)3 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/2/1

Prove that : $\frac{cosec θ - 1}{cosec θ + 1} + \frac{cosec θ + 1}{cosec θ - 1} = 2 sec θ$

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Answer: Proved.

LHS $= \frac{( cosec θ - 1 ) + ( cosec θ + 1 )}{( cosec θ + 1 ) ( cosec θ - 1 )}$
$= \frac{2 cosec θ}{cosec ^{2} θ - 1} = \frac{2 cosec θ}{c o t θ}$ (taking $θ$ acute)
$= 2 \times \frac{1}{s i n θ} \times \frac{s i n θ}{c o s θ} = \frac{2}{c o s θ} = 2 sec θ$ = RHS

Q263 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/2/2

If $tan θ + sin θ = m$ and $tan θ - sin θ = n$ , then prove that $m^{2} - n^{2} = 4 mn$ .

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Answer: Proved.

$m^{2} - n^{2} = (m + n) (m - n) = (2 tan θ) (2 sin θ) = 4 tan θ sin θ$
$mn = tan^{2} θ - sin^{2} θ = \frac{s i n ^{2} θ}{c o s ^{2} θ} - sin^{2} θ = sin^{2} θ (\frac{1 - c o s ^{2} θ}{c o s ^{2} θ})$
$= sin^{2} θ \cdot tan^{2} θ$
$mn = tan θ sin θ$ (taking $θ$ acute)
So $4 mn = 4 tan θ sin θ = m^{2} - n^{2}$

Q26 (OR) (OR)3 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/2/2

Prove that : $\frac{c o t A - 1}{2 - s e c ^{2} A} = \frac{c o t A}{1 + t a n A}$

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Answer: Proved.

$2 - sec^{2} A = 2 - (1 + tan^{2} A) = 1 - tan^{2} A = (1 - tan A) (1 + tan A)$
$cot A - 1 = \frac{1}{t a n A} - 1 = \frac{1 - t a n A}{t a n A}$
LHS $= \frac{1 - t a n A}{t a n A ( 1 - t a n A ) ( 1 + t a n A )} = \frac{1}{t a n A ( 1 + t a n A )}$
$= \frac{c o t A}{1 + t a n A}$ = RHS

Q26 (OR) (OR)3 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/2/3

If $cosec θ = x + \frac{1}{4 x}$ , prove that $cosec θ + cot θ = 2 x$ or $\frac{1}{2 x}$ .

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Answer: Proved.

$cot^{2} θ = cosec^{2} θ - 1 = x^{2} + \frac{1}{2} + \frac{1}{16 x ^{2}} - 1 = x^{2} - \frac{1}{2} + \frac{1}{16 x ^{2}}$
$= (x - \frac{1}{4 x})^{2}$ , so $cot θ = \pm (x - \frac{1}{4 x})$
If $cot θ = x - \frac{1}{4 x}$ : $cosec θ + cot θ = x + \frac{1}{4 x} + x - \frac{1}{4 x} = 2 x$
If $cot θ = - (x - \frac{1}{4 x})$ : $cosec θ + cot θ = x + \frac{1}{4 x} - x + \frac{1}{4 x} = \frac{1}{2 x}$
Hence $cosec θ + cot θ = 2 x$ or $\frac{1}{2 x}$

Q28 (OR) (OR)3 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/3/1

Prove that : $(\frac{1}{c o s A} - cos A) (\frac{1}{s i n A} - sin A) = \frac{1}{t a n A + c o t A}$

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Answer: Proved.

LHS $= \frac{1 - c o s ^{2} A}{c o s A} \times \frac{1 - s i n ^{2} A}{s i n A} = \frac{s i n ^{2} A}{c o s A} \times \frac{c o s ^{2} A}{s i n A} = sin A cos A$ .
RHS $= \frac{1}{\frac{s i n A}{c o s A} + \frac{c o s A}{s i n A}} = \frac{s i n A c o s A}{s i n ^{2} A + c o s ^{2} A} = sin A cos A$ .
LHS = RHS.

Also asked in: 2025 Standard 30/3/2, 2025 Standard 30/3/3

Q293 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/4/1

Prove that $\frac{c o s A + s i n A - 1}{c o s A - s i n A + 1} = cosec A - cot A$

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Answer: Proved.

Divide numerator and denominator by $sin A$ : LHS $= \frac{c o t A + 1 - cosec A}{c o t A - 1 + cosec A}$
Write $1 = cosec^{2} A - cot^{2} A$ in the numerator:
Numerator $= (cot A - cosec A) + (cosec A - cot A) (cosec A + cot A)$
$= (cosec A - cot A) (cosec A + cot A - 1)$
Denominator $= cosec A + cot A - 1$
LHS $= cosec A - cot A =$ RHS

Also asked in: 2025 Standard 30/4/2, 2025 Standard 30/4/3

Q29 (OR) (OR)3 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/4/1

If $cot θ + cos θ = p$ and $cot θ - cos θ = q$ ,
prove that $p^{2} - q^{2} = 4 pq$

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Answer: Proved.

$p^{2} - q^{2} = (p + q) (p - q) = 2 cot θ \times 2 cos θ = 4 cot θ cos θ$
$pq = cot^{2} θ - cos^{2} θ = \frac{c o s ^{2} θ}{s i n ^{2} θ} - cos^{2} θ = \frac{c o s ^{2} θ ( 1 - s i n ^{2} θ )}{s i n ^{2} θ} = cot^{2} θ cos^{2} θ$
$4 pq = 4 cot θ cos θ$
Hence $p^{2} - q^{2} = 4 pq$

Also asked in: 2025 Standard 30/4/2, 2025 Standard 30/4/3

Q303 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/5/1

Prove the following trigonometric identity :
$\frac{1 + cosec A}{cosec A} = \frac{c o s ^{2} A}{1 - s i n A}$

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Answer: Proved.

LHS $= \frac{1}{cosec A} + 1 = sin A + 1$
RHS $= \frac{1 - s i n ^{2} A}{1 - s i n A} = \frac{( 1 - s i n A ) ( 1 + s i n A )}{1 - s i n A} = 1 + sin A$
LHS = RHS. Proved.

Also asked in: 2025 Standard 30/5/2, 2025 Standard 30/5/3

Q30 (OR) (OR)3 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/5/1

Let $2 A + B$ and $A + 2 B$ be acute angles such that $sin (2 A + B) = \frac{3}{2}$ and $tan (A + 2 B) = 1$ . Find the value of $cot (4 A - 7 B)$ .

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Answer:

3

$sin (2 A + B) = \frac{3}{2} \Rightarrow 2 A + B = 6 0^{\circ}$
$tan (A + 2 B) = 1 \Rightarrow A + 2 B = 4 5^{\circ}$
Solving: $3 B = 2 (4 5^{\circ}) - 6 0^{\circ} = 3 0^{\circ} \Rightarrow B = 1 0^{\circ}$ , $A = 2 5^{\circ}$
$4 A - 7 B = 10 0^{\circ} - 7 0^{\circ} = 3 0^{\circ}$
$cot 3 0^{\circ} = 3$

Also asked in: 2025 Standard 30/5/2, 2025 Standard 30/5/3

Q303 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/6/1

Prove that : $\frac{c o s θ - 2 c o s ^{3} θ}{s i n θ - 2 s i n ^{3} θ} + cot θ = 0$ .

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Answer: Proved.

LHS first term $= \frac{c o s θ ( 1 - 2 c o s ^{2} θ )}{s i n θ ( 1 - 2 s i n ^{2} θ )}$
$1 - 2 cos^{2} θ = sin^{2} θ - cos^{2} θ$ and $1 - 2 sin^{2} θ = cos^{2} θ - sin^{2} θ$
So the term $= cot θ \cdot \frac{s i n ^{2} θ - c o s ^{2} θ}{c o s ^{2} θ - s i n ^{2} θ} = - cot θ$
LHS $= - cot θ + cot θ = 0 =$ RHS

Also asked in: 2025 Standard 30/6/2, 2025 Standard 30/6/3

Q30 (OR) (OR)3 marksShort AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/6/1

Given that $sin θ + cos θ = x$ , prove that $sin^{4} θ + cos^{4} θ = \frac{2 - ( x ^{2} - 1 ) ^{2}}{2}$ .

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Answer: Proved.

Squaring: $1 + 2 sin θ cos θ = x^{2} \Rightarrow sin θ cos θ = \frac{x ^{2} - 1}{2}$
$sin^{4} θ + cos^{4} θ = (sin^{2} θ + cos^{2} θ)^{2} - 2 sin^{2} θ cos^{2} θ = 1 - 2 (\frac{x ^{2} - 1}{2})^{2}$
$= 1 - \frac{( x ^{2} - 1 ) ^{2}}{2} = \frac{2 - ( x ^{2} - 1 ) ^{2}}{2}$

Also asked in: 2025 Standard 30/6/2, 2025 Standard 30/6/3

Q263 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Basic 430/1/2

Prove that : $\frac{s e c A - 1}{s e c A + 1} + \frac{s e c A + 1}{s e c A - 1} = 2 cosec A$

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Answer: Proved.

LHS $= \frac{( s e c A - 1 ) + ( s e c A + 1 )}{( s e c A + 1 ) ( s e c A - 1 )}$ (taking the common denominator).
$= \frac{2 s e c A}{s e c ^{2} A - 1} = \frac{2 s e c A}{t a n A}$ (for acute A).
$= 2 \times \frac{1}{c o s A} \times \frac{c o s A}{s i n A} = \frac{2}{s i n A}$ .
$= 2 cosec A =$ RHS.

Also asked in: 2025 Standard 30/3/1, 2025 Standard 30/3/2, 2025 Standard 30/3/3

Q313 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Basic 430/1/3

Prove that : $(cot θ - cosec θ)^{2} = \frac{1 - c o s θ}{1 + c o s θ}$ .

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Answer: Proved.

LHS $= (\frac{c o s θ}{s i n θ} - \frac{1}{s i n θ})^{2} = \frac{( 1 - c o s θ ) ^{2}}{s i n ^{2} θ}$ .
$= \frac{( 1 - c o s θ ) ^{2}}{1 - c o s ^{2} θ} = \frac{( 1 - c o s θ ) ^{2}}{( 1 - c o s θ ) ( 1 + c o s θ )}$ .
$= \frac{1 - c o s θ}{1 + c o s θ} =$ RHS.

Q303 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Basic 430/2/1

Prove that $\frac{t a n θ}{1 - c o t θ} + \frac{c o t θ}{1 - t a n θ} = 1 + sec θ cosec θ$

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Answer: Proved.

Let $t = tan θ$ , so $cot θ = \frac{1}{t}$ .
LHS $= \frac{t}{1 - \frac{1}{t}} + \frac{1/ t}{1 - t} = \frac{t ^{2}}{t - 1} - \frac{1}{t ( t - 1 )}$
$= \frac{t ^{3} - 1}{t ( t - 1 )} = \frac{( t - 1 ) ( t ^{2} + t + 1 )}{t ( t - 1 )} = \frac{t ^{2} + t + 1}{t} = 1 + t + \frac{1}{t}$
$t + \frac{1}{t} = \frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ} = \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n θ c o s θ} = sec θ cosec θ$
So LHS $= 1 + sec θ cosec θ =$ RHS.

Also asked in: 2024 Basic 430/2/2, 2024 Basic 430/2/3

Q313 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Basic 430/3/1

Prove that $(sin θ + cosec θ)^{2} + (cos θ + sec θ)^{2} = 7 + tan^{2} θ + cot^{2} θ$ .

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Answer: Proved.

LHS $= sin^{2} θ + 2 sin θ cosec θ + cosec^{2} θ + cos^{2} θ + 2 cos θ sec θ + sec^{2} θ$ .
$= (sin^{2} θ + cos^{2} θ) + 2 + 2 + cosec^{2} θ + sec^{2} θ$ .
$= 1 + 4 + (1 + cot^{2} θ) + (1 + tan^{2} θ)$ .
$= 7 + tan^{2} θ + cot^{2} θ =$ RHS.

Also asked in: 2024 Basic 430/3/2, 2024 Basic 430/3/3, 2024 Basic 430/5/1

Q31 (OR) (OR)3 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Basic 430/3/1

If $cos A = \frac{5}{13}$ , then verify that $\frac{c o s A}{1 - t a n A} + \frac{s i n A}{1 - c o t A} = cos A + sin A$ .

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Answer: Verified; both sides equal

\frac{17}{13}

$sin A = 1 - \frac{25}{169} = \frac{12}{13}$ , so $tan A = \frac{12}{5}$ and $cot A = \frac{5}{12}$ .
$\frac{c o s A}{1 - t a n A} = \frac{5/13}{- 7/5} = - \frac{25}{91}$ .
$\frac{s i n A}{1 - c o t A} = \frac{12/13}{7/12} = \frac{144}{91}$ .
LHS $= \frac{144 - 25}{91} = \frac{119}{91} = \frac{17}{13}$ .
RHS $= \frac{5}{13} + \frac{12}{13} = \frac{17}{13}$ . Hence verified.

Also asked in: 2024 Basic 430/3/2, 2024 Basic 430/3/3

Q313 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Basic 430/4/1

Prove that :
$(tan A + sec A)^{2} + (tan A - sec A)^{2} = 2 (\frac{1 + s i n ^{2} A}{1 - s i n ^{2} A})$

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Answer: Proved.

LHS $= 2 tan^{2} A + 2 sec^{2} A$ (the cross terms cancel).
$= 2 (\frac{s i n ^{2} A}{c o s ^{2} A} + \frac{1}{c o s ^{2} A}) = 2 (\frac{1 + s i n ^{2} A}{c o s ^{2} A})$ .
Since $cos^{2} A = 1 - sin^{2} A$ , LHS $= 2 (\frac{1 + s i n ^{2} A}{1 - s i n ^{2} A}) =$ RHS.

Q263 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Basic 430/4/2

Prove that :
$(cosec A - cot A)^{2} = \frac{1 - c o s A}{1 + c o s A}$

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Answer: Proved.

LHS $= (\frac{1}{s i n A} - \frac{c o s A}{s i n A})^{2} = \frac{( 1 - c o s A ) ^{2}}{s i n ^{2} A}$ .
$= \frac{( 1 - c o s A ) ^{2}}{1 - c o s ^{2} A} = \frac{( 1 - c o s A ) ^{2}}{( 1 - c o s A ) ( 1 + c o s A )}$ .
$= \frac{1 - c o s A}{1 + c o s A} =$ RHS.

Q303 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Basic 430/5/2

Prove that : $(cosec θ - sin θ) (sec θ - cos θ) = \frac{1}{t a n θ + c o t θ}$ .

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Answer: Proved.

LHS $= (\frac{1}{s i n θ} - sin θ) (\frac{1}{c o s θ} - cos θ) = \frac{1 - s i n ^{2} θ}{s i n θ} \cdot \frac{1 - c o s ^{2} θ}{c o s θ}$ .
$= \frac{c o s ^{2} θ}{s i n θ} \cdot \frac{s i n ^{2} θ}{c o s θ} = sin θ cos θ$ .
RHS $= \frac{1}{\frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ}} = \frac{s i n θ c o s θ}{s i n ^{2} θ + c o s ^{2} θ} = sin θ cos θ$ .
Hence LHS = RHS.

Q303 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Basic 430/5/3

Prove that $\frac{1 + s i n A}{1 - s i n A} = sec A + tan A$ .

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Answer: Proved.

Multiply numerator and denominator inside the root by $(1 + sin A)$ .
LHS $= \frac{( 1 + s i n A ) ^{2}}{1 - s i n ^{2} A} = \frac{( 1 + s i n A ) ^{2}}{c o s ^{2} A}$ .
$= \frac{1 + s i n A}{c o s A}$ (A acute, so $cos A > 0$ ).
$= \frac{1}{c o s A} + \frac{s i n A}{c o s A} = sec A + tan A$ = RHS.

Also asked in: 2026 Standard 30/4/1, 2026 Standard 30/4/2

Q293 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/2/1

Prove that $\frac{1 + s e c θ - t a n θ}{1 + s e c θ + t a n θ} = \frac{1 - s i n θ}{c o s θ}$ .

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Answer: Proved.

Use $1 = sec^{2} θ - tan^{2} θ = (sec θ - tan θ) (sec θ + tan θ)$ in the numerator.
Numerator $= (sec θ - tan θ) + (sec θ - tan θ) (sec θ + tan θ) = (sec θ - tan θ) (1 + sec θ + tan θ)$
So LHS $= sec θ - tan θ = \frac{1}{c o s θ} - \frac{s i n θ}{c o s θ} = \frac{1 - s i n θ}{c o s θ}$ = RHS.

Q313 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/2/2

Prove that $sin^{6} θ + cos^{6} θ = 1 - 3 sin^{2} θ cos^{2} θ$ .

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Answer: Proved.

Use $a^{3} + b^{3} = (a + b)^{3} - 3 ab (a + b)$ with $a = sin^{2} θ$ , $b = cos^{2} θ$ .
LHS $= (sin^{2} θ + cos^{2} θ)^{3} - 3 sin^{2} θ cos^{2} θ (sin^{2} θ + cos^{2} θ)$
$= 1^{3} - 3 sin^{2} θ cos^{2} θ \times 1 = 1 - 3 sin^{2} θ cos^{2} θ$ = RHS

Q263 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/2/3

Prove that $\frac{cosec ^{2} θ - s e c ^{2} θ}{cosec ^{2} θ + s e c ^{2} θ} = \frac{3}{4}$ , if $tan θ = \frac{1}{7}$

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Answer: Proved.

$tan θ = \frac{1}{7}$ , so $tan^{2} θ = \frac{1}{7}$ and $cot^{2} θ = 7$ .
$cosec^{2} θ = 1 + cot^{2} θ = 8$ , $sec^{2} θ = 1 + tan^{2} θ = \frac{8}{7}$
LHS $= \frac{8 - \frac{8}{7}}{8 + \frac{8}{7}} = \frac{\frac{48}{7}}{\frac{64}{7}} = \frac{48}{64} = \frac{3}{4}$ = RHS

Q303 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/3/1

Prove that $\frac{s i n θ - 2 s i n ^{3} θ}{2 c o s ^{3} θ - c o s θ} = tan θ$ .

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Answer: Proved.

LHS = $\frac{s i n θ ( 1 - 2 s i n ^{2} θ )}{c o s θ ( 2 c o s ^{2} θ - 1 )}$ .
$1 - 2 sin^{2} θ = 1 - 2 (1 - cos^{2} θ) = 2 cos^{2} θ - 1$ .
So LHS = $\frac{s i n θ ( 2 c o s ^{2} θ - 1 )}{c o s θ ( 2 c o s ^{2} θ - 1 )} = \frac{s i n θ}{c o s θ} = tan θ$ = RHS.

Q273 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/3/2

Prove that $\frac{s i n A + c o s A}{s i n A - c o s A} + \frac{s i n A - c o s A}{s i n A + c o s A} = \frac{2}{2 s i n ^{2} A - 1}$

Show answer & solution

Answer: Proved.

LHS = $\frac{( s i n A + c o s A ) ^{2} + ( s i n A - c o s A ) ^{2}}{( s i n A - c o s A ) ( s i n A + c o s A )}$ .
Numerator = $2 (sin^{2} A + cos^{2} A) = 2$ .
Denominator = $sin^{2} A - cos^{2} A = sin^{2} A - (1 - sin^{2} A) = 2 sin^{2} A - 1$ .
So LHS = $\frac{2}{2 s i n ^{2} A - 1}$ = RHS.

Also asked in: 2025 Standard 30/1/1, 2025 Standard 30/1/2, 2025 Standard 30/1/3

Q313 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/3/3

Prove that :
$(cosec θ - sin θ) (sec θ - cos θ) (tan θ + cot θ) = 1$

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Answer: Proved.

$cosec θ - sin θ = \frac{1 - s i n ^{2} θ}{s i n θ} = \frac{c o s ^{2} θ}{s i n θ}$ .
$sec θ - cos θ = \frac{1 - c o s ^{2} θ}{c o s θ} = \frac{s i n ^{2} θ}{c o s θ}$ .
$tan θ + cot θ = \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n θ c o s θ} = \frac{1}{s i n θ c o s θ}$ .
Product = $\frac{c o s ^{2} θ}{s i n θ} \times \frac{s i n ^{2} θ}{c o s θ} \times \frac{1}{s i n θ c o s θ} = 1$ = RHS.

Q263 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/4/1

Prove that $\frac{s i n θ - c o s θ + 1}{s i n θ + c o s θ - 1} = \frac{1}{s e c θ - t a n θ}$

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Answer: Proved.

Divide numerator and denominator of the LHS by $cos θ$ : LHS = $\frac{t a n θ - 1 + s e c θ}{t a n θ + 1 - s e c θ}$ .
Write the 1 in the numerator as $sec^{2} θ - tan^{2} θ$ : numerator = $(sec θ + tan θ) - (sec θ + tan θ) (sec θ - tan θ)$ .
Numerator = $(sec θ + tan θ) (1 - sec θ + tan θ)$ .
The bracket equals the denominator $tan θ + 1 - sec θ$ , so LHS = $sec θ + tan θ$ .
Since $(sec θ + tan θ) (sec θ - tan θ) = 1$ , $sec θ + tan θ = \frac{1}{s e c θ - t a n θ}$ = RHS.

Also asked in: 2026 Standard 30/4/1

Q313 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/4/3

Prove that $sec^{2} θ + cosec^{2} θ = tan θ + cot θ$ .

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Answer: Proved.

$sec^{2} θ + cosec^{2} θ = \frac{1}{c o s ^{2} θ} + \frac{1}{s i n ^{2} θ} = \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n ^{2} θ c o s ^{2} θ} = \frac{1}{s i n ^{2} θ c o s ^{2} θ}$ .
So LHS = $\frac{1}{s i n θ c o s θ}$ (for acute $θ$ ).
RHS = $\frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ} = \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n θ c o s θ} = \frac{1}{s i n θ c o s θ}$ .
LHS = RHS. Hence proved.

Also asked in: 2025 Standard 30/2/3

Q293 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/5/1

Prove that :
$\frac{t a n A}{1 - c o t A} + \frac{c o t A}{1 - t a n A} = 1 + sec A cosec A$

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Answer: Proved.

Write $t = tan A$ , so $cot A = \frac{1}{t}$ .
LHS $= \frac{t}{1 - \frac{1}{t}} + \frac{\frac{1}{t}}{1 - t} = \frac{t ^{2}}{t - 1} - \frac{1}{t ( t - 1 )} = \frac{t ^{3} - 1}{t ( t - 1 )}$ .
$= \frac{( t - 1 ) ( t ^{2} + t + 1 )}{t ( t - 1 )} = \frac{t ^{2} + t + 1}{t} = t + 1 + \frac{1}{t}$ .
$= 1 + tan A + cot A = 1 + \frac{s i n ^{2} A + c o s ^{2} A}{s i n A c o s A} = 1 + \frac{1}{s i n A c o s A}$ .
$= 1 + sec A cosec A$ = RHS.

Q303 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/5/2

Prove that :
$\frac{s i n θ}{1 + c o s θ} + \frac{1 + c o s θ}{s i n θ} = 2 cosec θ$

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Answer: Proved.

LHS $= \frac{s i n ^{2} θ + ( 1 + c o s θ ) ^{2}}{s i n θ ( 1 + c o s θ )}$ .
$= \frac{s i n ^{2} θ + 1 + 2 c o s θ + c o s ^{2} θ}{s i n θ ( 1 + c o s θ )} = \frac{2 + 2 c o s θ}{s i n θ ( 1 + c o s θ )}$ .
$= \frac{2 ( 1 + c o s θ )}{s i n θ ( 1 + c o s θ )} = \frac{2}{s i n θ} = 2 cosec θ$ = RHS.

Q263 marksShort AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/5/3

Prove that :
$\frac{t a n θ - c o t θ}{s i n θ c o s θ} = sec^{2} θ - cosec^{2} θ$

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Answer: Proved.

LHS $= \frac{t a n θ}{s i n θ c o s θ} - \frac{c o t θ}{s i n θ c o s θ}$ .
$\frac{t a n θ}{s i n θ c o s θ} = \frac{s i n θ}{c o s θ} \times \frac{1}{s i n θ c o s θ} = \frac{1}{c o s ^{2} θ} = sec^{2} θ$ .
$\frac{c o t θ}{s i n θ c o s θ} = \frac{c o s θ}{s i n θ} \times \frac{1}{s i n θ c o s θ} = \frac{1}{s i n ^{2} θ} = cosec^{2} θ$ .
So LHS $= sec^{2} θ - cosec^{2} θ$ = RHS.

Q263 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/1/1

Prove that
$sec θ (1 - sin θ) (sec θ + tan θ) = 1$

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Answer: Proved.

LHS $= \frac{1}{c o s θ} (1 - sin θ) (\frac{1}{c o s θ} + \frac{s i n θ}{c o s θ})$
$= \frac{( 1 - s i n θ ) ( 1 + s i n θ )}{c o s ^{2} θ}$
$= \frac{1 - s i n ^{2} θ}{c o s ^{2} θ} = \frac{c o s ^{2} θ}{c o s ^{2} θ} = 1 =$ RHS

Also asked in: 2023 Basic 430/1/2

Q26 (OR) (OR)3 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/1/1

Prove that
$\frac{1 + s e c θ}{s e c θ} = \frac{s i n ^{2} θ}{1 - c o s θ}$

Show answer & solution

Answer: Proved.

LHS $= \frac{1 + \frac{1}{c o s θ}}{\frac{1}{c o s θ}} = cos θ + 1$
RHS $= \frac{1 - c o s ^{2} θ}{1 - c o s θ} = \frac{( 1 - c o s θ ) ( 1 + c o s θ )}{1 - c o s θ} = 1 + cos θ$
LHS = RHS

Also asked in: 2024 Basic 430/4/3, 2023 Basic 430/1/2, 2023 Basic 430/4/3

Q313 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/1/3

Prove that $\frac{c o s A}{1 + s i n A} + \frac{1 + s i n A}{c o s A} = 2 sec A$ .

Show answer & solution

Answer: Proved.

LHS $= \frac{c o s ^{2} A + ( 1 + s i n A ) ^{2}}{( 1 + s i n A ) c o s A}$
$= \frac{c o s ^{2} A + 1 + 2 s i n A + s i n ^{2} A}{( 1 + s i n A ) c o s A}$
$= \frac{2 + 2 s i n A}{( 1 + s i n A ) c o s A}$ (since $sin^{2} A + cos^{2} A = 1$ )
$= \frac{2 ( 1 + s i n A )}{( 1 + s i n A ) c o s A} = \frac{2}{c o s A} = 2 sec A =$ RHS

Q31 (OR) (OR)3 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/1/3

Prove that $(sin A + cosec A)^{2} + (cos A + sec A)^{2} = 7 + tan^{2} A + cot^{2} A$ .

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Answer: Proved.

LHS $= sin^{2} A + 2 sin A cosec A + cosec^{2} A + cos^{2} A + 2 cos A sec A + sec^{2} A$
$= (sin^{2} A + cos^{2} A) + 2 + 2 + cosec^{2} A + sec^{2} A$ (since $sin A cosec A = cos A sec A = 1$ )
$= 1 + 4 + (1 + cot^{2} A) + (1 + tan^{2} A)$
$= 7 + tan^{2} A + cot^{2} A =$ RHS

Q293 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/2/1

Prove that :
$\frac{1 - c o s θ}{1 + c o s θ} = (cosec θ - cot θ)^{2}$

Show answer & solution

Answer: Proved.

RHS $= (\frac{1}{s i n θ} - \frac{c o s θ}{s i n θ})^{2} = \frac{( 1 - c o s θ ) ^{2}}{s i n ^{2} θ}$
$= \frac{( 1 - c o s θ ) ^{2}}{1 - c o s ^{2} θ} = \frac{( 1 - c o s θ ) ^{2}}{( 1 - c o s θ ) ( 1 + c o s θ )}$
$= \frac{1 - c o s θ}{1 + c o s θ}$ = LHS

Also asked in: 2023 Basic 430/2/2, 2023 Basic 430/2/3

Q29 (OR) (OR)3 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/2/1

Prove that :
$(1 + \frac{1}{t a n ^{2} A}) (1 + \frac{1}{c o t ^{2} A}) = \frac{1}{s i n ^{2} A - s i n ^{4} A}$

Show answer & solution

Answer: Proved.

LHS $= (1 + cot^{2} A) (1 + tan^{2} A) = cosec^{2} A \cdot sec^{2} A$
$= \frac{1}{s i n ^{2} A c o s ^{2} A} = \frac{1}{s i n ^{2} A ( 1 - s i n ^{2} A )}$
$= \frac{1}{s i n ^{2} A - s i n ^{4} A}$ = RHS

Also asked in: 2023 Basic 430/2/2, 2023 Basic 430/2/3

Q273 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/4/1

Prove that $\frac{1 + t a n ^{2} A}{1 + c o t ^{2} A} = sec^{2} A - 1$

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Answer: Proved.

LHS $= \frac{1 + t a n ^{2} A}{1 + c o t ^{2} A} = \frac{s e c ^{2} A}{c s c ^{2} A}$ .
$= \frac{1/ c o s ^{2} A}{1/ s i n ^{2} A} = \frac{s i n ^{2} A}{c o s ^{2} A} = tan^{2} A$ .
$= sec^{2} A - 1$ = RHS.

Also asked in: 2023 Basic 430/4/2

Q303 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/5/1

Prove that :
$(cosec A - sin A) (sec A - cos A) = \frac{1}{t a n A + c o t A}$

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Answer: Proved.

LHS $= (\frac{1}{s i n A} - sin A) (\frac{1}{c o s A} - cos A) = \frac{1 - s i n ^{2} A}{s i n A} \cdot \frac{1 - c o s ^{2} A}{c o s A}$ .
$= \frac{c o s ^{2} A}{s i n A} \cdot \frac{s i n ^{2} A}{c o s A} = sin A cos A$ .
RHS $= \frac{1}{\frac{s i n A}{c o s A} + \frac{c o s A}{s i n A}} = \frac{s i n A c o s A}{s i n ^{2} A + c o s ^{2} A} = sin A cos A$ .
LHS = RHS. Hence proved.

Also asked in: 2026 Basic 430/4/1, 2026 Basic 430/4/3

Q263 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/5/2

Prove that :
$\frac{s i n θ + c o s θ}{s i n θ - c o s θ} + \frac{s i n θ - c o s θ}{s i n θ + c o s θ} = \frac{2 s e c ^{2} θ}{t a n ^{2} θ - 1}$

Show answer & solution

Answer: Proved.

LHS $= \frac{( s i n θ + c o s θ ) ^{2} + ( s i n θ - c o s θ ) ^{2}}{s i n ^{2} θ - c o s ^{2} θ}$ .
Numerator $= 2 (sin^{2} θ + cos^{2} θ) = 2$ , so LHS $= \frac{2}{s i n ^{2} θ - c o s ^{2} θ}$ .
Divide numerator and denominator by $cos^{2} θ$ : LHS $= \frac{2 s e c ^{2} θ}{t a n ^{2} θ - 1}$ = RHS.
Hence proved.

Q293 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/5/3

Prove that :
$\frac{s e c θ - 1}{s e c θ + 1} + \frac{s e c θ + 1}{s e c θ - 1} = 2 cosec θ$

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Answer: Proved.

Take $θ$ acute, so all the quantities are positive.
$\frac{s e c θ - 1}{s e c θ + 1} = \frac{s e c θ - 1}{s e c ^{2} θ - 1} = \frac{s e c θ - 1}{t a n θ}$ and $\frac{s e c θ + 1}{s e c θ - 1} = \frac{s e c θ + 1}{t a n θ}$ .
LHS $= \frac{( s e c θ - 1 ) + ( s e c θ + 1 )}{t a n θ} = \frac{2 s e c θ}{t a n θ}$ .
$= 2 \times \frac{1}{c o s θ} \times \frac{c o s θ}{s i n θ} = \frac{2}{s i n θ} = 2 cosec θ$ = RHS.
Hence proved.

Q313 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/6/1

Prove that $\frac{c o t A - c o s A}{c o t A + c o s A} = \frac{c o s ^{2} A}{( 1 + s i n A ) ^{2}}$

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Answer: Proved.

LHS $= \frac{\frac{c o s A}{s i n A} - c o s A}{\frac{c o s A}{s i n A} + c o s A} = \frac{c o s A ( \frac{1}{s i n A} - 1 )}{c o s A ( \frac{1}{s i n A} + 1 )} = \frac{1 - s i n A}{1 + s i n A}$
Multiply numerator and denominator by $(1 + sin A)$ : $= \frac{1 - s i n ^{2} A}{( 1 + s i n A ) ^{2}} = \frac{c o s ^{2} A}{( 1 + s i n A ) ^{2}}$ = RHS.

Also asked in: 2023 Basic 430/6/2, 2023 Basic 430/6/3

Q31 (OR) (OR)3 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/6/1

Prove that $(sec θ + tan θ) (1 - sin θ) = cos θ$

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Answer: Proved.

LHS $= (\frac{1}{c o s θ} + \frac{s i n θ}{c o s θ}) (1 - sin θ) = \frac{( 1 + s i n θ ) ( 1 - s i n θ )}{c o s θ}$
$= \frac{1 - s i n ^{2} θ}{c o s θ} = \frac{c o s ^{2} θ}{c o s θ} = cos θ$ = RHS.

Also asked in: 2023 Basic 430/6/2, 2023 Basic 430/6/3

Q303 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/1/1

Prove that : $\frac{t a n θ + s e c θ - 1}{t a n θ - s e c θ + 1} = \frac{1 + s i n θ}{c o s θ}$

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Answer: Proved.

Use $sec^{2} θ - tan^{2} θ = 1$ , i.e. $1 = (sec θ - tan θ) (sec θ + tan θ)$ .
Numerator: $tan θ + sec θ - (sec^{2} θ - tan^{2} θ) = (sec θ + tan θ) [1 - sec θ + tan θ]$ .
So LHS $= \frac{( s e c θ + t a n θ ) ( t a n θ - s e c θ + 1 )}{t a n θ - s e c θ + 1} = sec θ + tan θ$ .
$= \frac{1}{c o s θ} + \frac{s i n θ}{c o s θ} = \frac{1 + s i n θ}{c o s θ}$ = RHS.

Q313 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/1/2

Prove that $(cosec A - sin A) (sec A - cos A) = \frac{1}{c o t A + t a n A}$ .

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Answer: Proved.

LHS $= (\frac{1}{s i n A} - sin A) (\frac{1}{c o s A} - cos A) = \frac{c o s ^{2} A}{s i n A} \cdot \frac{s i n ^{2} A}{c o s A} = sin A cos A$ .
RHS $= \frac{1}{\frac{c o s A}{s i n A} + \frac{s i n A}{c o s A}} = \frac{s i n A c o s A}{c o s ^{2} A + s i n ^{2} A} = sin A cos A$ .
LHS = RHS.

Q273 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/1/3

Prove that : $2 (sin^{6} θ + cos^{6} θ) - 3 (sin^{4} θ + cos^{4} θ) + 1 = 0$ .

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Answer: Proved.

$sin^{6} θ + cos^{6} θ = (sin^{2} θ + cos^{2} θ)^{3} - 3 sin^{2} θ cos^{2} θ (sin^{2} θ + cos^{2} θ) = 1 - 3 sin^{2} θ cos^{2} θ$ .
$sin^{4} θ + cos^{4} θ = (sin^{2} θ + cos^{2} θ)^{2} - 2 sin^{2} θ cos^{2} θ = 1 - 2 sin^{2} θ cos^{2} θ$ .
LHS $= 2 - 6 sin^{2} θ cos^{2} θ - 3 + 6 sin^{2} θ cos^{2} θ + 1 = 0$ = RHS.

Q273 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/2/1

Prove that :
$(\frac{1}{c o s θ} - cos θ) (\frac{1}{s i n θ} - sin θ) = \frac{1}{t a n θ + c o t θ}$ .

Show answer & solution

Answer: Proved.

LHS $= \frac{1 - c o s ^{2} θ}{c o s θ} \times \frac{1 - s i n ^{2} θ}{s i n θ} = \frac{s i n ^{2} θ}{c o s θ} \times \frac{c o s ^{2} θ}{s i n θ} = sin θ cos θ$
RHS $= \frac{1}{\frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ}} = \frac{s i n θ c o s θ}{s i n ^{2} θ + c o s ^{2} θ} = sin θ cos θ$
LHS = RHS. Hence proved.

Q263 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/2/2

Prove that :
$\frac{s i n θ}{1 + c o s θ} + \frac{1 + c o s θ}{s i n θ} = 2 cosec θ$

Show answer & solution

Answer: Proved.

LHS $= \frac{s i n ^{2} θ + ( 1 + c o s θ ) ^{2}}{( 1 + c o s θ ) s i n θ}$
$= \frac{s i n ^{2} θ + c o s ^{2} θ + 1 + 2 c o s θ}{( 1 + c o s θ ) s i n θ} = \frac{2 + 2 c o s θ}{( 1 + c o s θ ) s i n θ}$
$= \frac{2 ( 1 + c o s θ )}{( 1 + c o s θ ) s i n θ} = \frac{2}{s i n θ} = 2 cosec θ$ = RHS

Also asked in: 2025 Basic 430/6/3

Q263 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/2/3

Prove that :
$\frac{c o s ^{2} θ}{1 - t a n θ} + \frac{s i n ^{3} θ}{s i n θ - c o s θ} = 1 + sin θ cos θ$

Show answer & solution

Answer: Proved.

$\frac{c o s ^{2} θ}{1 - t a n θ} = \frac{c o s ^{2} θ}{\frac{c o s θ - s i n θ}{c o s θ}} = \frac{c o s ^{3} θ}{c o s θ - s i n θ}$
LHS $= \frac{c o s ^{3} θ}{c o s θ - s i n θ} - \frac{s i n ^{3} θ}{c o s θ - s i n θ} = \frac{c o s ^{3} θ - s i n ^{3} θ}{c o s θ - s i n θ}$
$= \frac{( c o s θ - s i n θ ) ( c o s ^{2} θ + c o s θ s i n θ + s i n ^{2} θ )}{c o s θ - s i n θ}$
$= 1 + sin θ cos θ$ = RHS

Q293 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/4/1

Prove that $\frac{s i n A - 2 s i n ^{3} A}{2 c o s ^{3} A - c o s A} = tan A$

Show answer & solution

Answer: Proved.

LHS $= \frac{s i n A ( 1 - 2 s i n ^{2} A )}{c o s A ( 2 c o s ^{2} A - 1 )}$
$1 - 2 sin^{2} A = 1 - 2 (1 - cos^{2} A) = 2 cos^{2} A - 1$
So LHS $= \frac{s i n A ( 2 c o s ^{2} A - 1 )}{c o s A ( 2 c o s ^{2} A - 1 )} = \frac{s i n A}{c o s A} = tan A$ = RHS

Also asked in: 2023 Standard 30/4/2, 2023 Standard 30/4/3

Q29 (OR) (OR)3 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/4/1

Prove that $sec A (1 - sin A) (sec A + tan A) = 1$ .

Show answer & solution

Answer: Proved.

LHS $= \frac{1}{c o s A} (1 - sin A) \cdot \frac{1 + s i n A}{c o s A}$
$= \frac{1 - s i n ^{2} A}{c o s ^{2} A} = \frac{c o s ^{2} A}{c o s ^{2} A} = 1$ = RHS

Also asked in: 2023 Standard 30/4/2, 2023 Standard 30/4/3

Q303 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/5/1

Prove that :
$\frac{t a n θ}{1 - c o t θ} + \frac{c o t θ}{1 - t a n θ} = 1 + sec θ cosec θ$

Show answer & solution

Answer: Proved.

Let $t = tan θ$ , so $cot θ = \frac{1}{t}$ .
LHS $= \frac{t}{1 - \frac{1}{t}} + \frac{\frac{1}{t}}{1 - t} = \frac{t ^{2}}{t - 1} - \frac{1}{t ( t - 1 )} = \frac{t ^{3} - 1}{t ( t - 1 )}$ .
$t^{3} - 1 = (t - 1) (t^{2} + t + 1)$ , so LHS $= \frac{t ^{2} + t + 1}{t} = t + 1 + \frac{1}{t}$ .
$tan θ + cot θ = \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n θ c o s θ} = sec θ cosec θ$ .
So LHS $= 1 + sec θ cosec θ$ = RHS.

Also asked in: 2024 Standard 30/1/1, 2024 Standard 30/1/2, 2024 Standard 30/1/3, 2023 Standard 30/5/2

Q283 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/5/3

Prove that :
$\frac{t a n A}{1 + s e c A} - \frac{t a n A}{1 - s e c A} = 2 cosec A$

Show answer & solution

Answer: Proved.

LHS $= tan A \cdot \frac{( 1 - s e c A ) - ( 1 + s e c A )}{( 1 + s e c A ) ( 1 - s e c A )} = tan A \cdot \frac{- 2 s e c A}{1 - s e c ^{2} A}$ .
$1 - sec^{2} A = - tan^{2} A$ , so LHS $= \frac{2 s e c A}{t a n A}$ .
$\frac{s e c A}{t a n A} = \frac{1}{c o s A} \times \frac{c o s A}{s i n A} = \frac{1}{s i n A}$ .
LHS $= 2 cosec A$ = RHS.

Also asked in: 2026 Standard 30/1/1, 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q303 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/6/1

Prove that $\frac{1 + s e c A}{s e c A} = \frac{s i n ^{2} A}{1 - c o s A}$ .

Show answer & solution

Answer: Proved.

LHS $= \frac{1 + \frac{1}{c o s A}}{\frac{1}{c o s A}} = cos A + 1$
RHS $= \frac{1 - c o s ^{2} A}{1 - c o s A} = \frac{( 1 - c o s A ) ( 1 + c o s A )}{1 - c o s A} = 1 + cos A$
LHS = RHS.

Q273 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/6/2

If $sin θ + cos θ = p$ and $sec θ + cosec θ = q$ , then prove that $q (p^{2} - 1) = 2 p$ .

Show answer & solution

Answer: Proved.

$p^{2} = sin^{2} θ + cos^{2} θ + 2 sin θ cos θ = 1 + 2 sin θ cos θ$ , so $p^{2} - 1 = 2 sin θ cos θ$ .
$q = \frac{1}{c o s θ} + \frac{1}{s i n θ} = \frac{s i n θ + c o s θ}{s i n θ c o s θ} = \frac{p}{s i n θ c o s θ}$
$q (p^{2} - 1) = \frac{p}{s i n θ c o s θ} \times 2 sin θ cos θ = 2 p$

Also asked in: 2024 Standard 30/4/2

Q263 marksShort AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/6/3

Prove that $(sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ$ .

Show answer & solution

Answer: Proved.

$tan θ + cot θ = \frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ} = \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n θ c o s θ} = \frac{1}{s i n θ c o s θ}$
LHS $= \frac{s i n θ + c o s θ}{s i n θ c o s θ} = \frac{1}{c o s θ} + \frac{1}{s i n θ}$
$= sec θ + cosec θ$ = RHS

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