Introduction to Trigonometry: CBSE Class 10 Previous Year Questions

314 different questions from Introduction to Trigonometry (NCERT Chapter 8) asked in CBSE Class 10 Maths board exams 2022–2026. 12 of them came up in more than one year. Pick a mark group to practise, each with answers and step-by-step solutions.

1 mark questions139 questions 2 marks questions88 questions 3 marks questions86 questions 4 marks questions1 questions

Most asked Introduction to Trigonometry questions

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

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

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Answer: Proved.

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

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

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

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

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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

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

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

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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

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

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

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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 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

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 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

Q252 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/4/1

Evaluate $\frac{5 c o s ^{2} 6 0 ^{\circ} + 4 s e c ^{2} 3 0 ^{\circ} - t a n ^{2} 4 5 ^{\circ}}{s i n ^{2} 3 0 ^{\circ} + c o s ^{2} 3 0 ^{\circ}}$

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Answer:

\frac{67}{12}

$cos 6 0^{\circ} = \frac{1}{2}$ , $sec 3 0^{\circ} = \frac{2}{3}$ , $tan 4 5^{\circ} = 1$
Numerator $= 5 \cdot \frac{1}{4} + 4 \cdot \frac{4}{3} - 1 = \frac{5}{4} + \frac{16}{3} - 1 = \frac{15 + 64 - 12}{12} = \frac{67}{12}$
Denominator $= sin^{2} 3 0^{\circ} + cos^{2} 3 0^{\circ} = 1$
Value $= \frac{67}{12}$

Also asked in: 2026 Standard 30/2/2

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