CBSE Maths › Class 10 PYQs › Introduction to Trigonometry

Introduction to Trigonometry: 2 marks Questions (CBSE Class 10)

88 different 2 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)

Q212 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Basic 430/4/1

If $sin (A - B) = \frac{1}{2}$ and $tan (A + B) = 3$ , $0^{\circ} \leq A + B < 9 0^{\circ}$ , A > B then find the values of A and B.

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Answer: A =

4 5^{\circ}

, B =

1 5^{\circ}

$sin (A - B) = \frac{1}{2} \Rightarrow A - B = 3 0^{\circ}$ .
$tan (A + B) = 3 \Rightarrow A + B = 6 0^{\circ}$ .
Adding: $2 A = 9 0^{\circ}$ , so $A = 4 5^{\circ}$ .
$B = 6 0^{\circ} - 4 5^{\circ} = 1 5^{\circ}$ .

Q212 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Basic 430/4/2

If $sin 2 A = \frac{3}{2}$ and $2 tan B + 1 = 3$ , then find the value of (A + B).

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Answer: A + B =

7 5^{\circ}

$sin 2 A = \frac{3}{2} \Rightarrow 2 A = 6 0^{\circ}$ , so $A = 3 0^{\circ}$ .
$2 tan B + 1 = 3 \Rightarrow tan B = 1$ , so $B = 4 5^{\circ}$ .
$A + B = 7 5^{\circ}$ .

Q212 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Basic 430/4/3

If $sin A = \frac{1}{2}$ and $tan B = 3$ , then verify that cos (A + B) = cos A cos B – sin A sin B.

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Answer: Verified: both sides equal 0 (A =

3 0^{\circ}

, B =

6 0^{\circ}

$sin A = \frac{1}{2} \Rightarrow A = 3 0^{\circ}$ ; $tan B = 3 \Rightarrow B = 6 0^{\circ}$ .
LHS = $cos (3 0^{\circ} + 6 0^{\circ}) = cos 9 0^{\circ} = 0$ .
RHS = $cos 3 0^{\circ} cos 6 0^{\circ} - sin 3 0^{\circ} sin 6 0^{\circ} = \frac{3}{2} \cdot \frac{1}{2} - \frac{1}{2} \cdot \frac{3}{2} = 0$ .
LHS = RHS. Hence verified.

Q212 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Basic 430/5/1

If $sec A = 2$ and $tan B = 3$ , then find the value of $2 sin A cos B$ .

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Answer:

\frac{1}{2}

(i.e.

\frac{2}{2}

)

$sec A = 2$ gives $A = 4 5^{\circ}$ ; $tan B = 3$ gives $B = 6 0^{\circ}$ .
$2 sin A cos B = 2 \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{2} = \frac{2}{2}$

Also asked in: 2026 Basic 430/5/3

Q21 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Basic 430/5/1

Evaluate : $\frac{4 c o s ^{3} 6 0 ^{\circ} + cosec 3 0 ^{\circ}}{t a n ^{2} 3 0 ^{\circ}}$

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Answer:

\frac{15}{2}

$4 cos^{3} 6 0^{\circ} = 4 \times \frac{1}{8} = \frac{1}{2}$ , $cosec 3 0^{\circ} = 2$ , $tan^{2} 3 0^{\circ} = \frac{1}{3}$
Value $= \frac{\frac{1}{2} + 2}{\frac{1}{3}} = \frac{5}{2} \times 3 = \frac{15}{2}$

Also asked in: 2026 Basic 430/5/3

Q232 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Basic 430/5/2

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

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Answer:

\frac{9}{16}

$5 sin^{2} 4 5^{\circ} = 5 \times \frac{1}{2} = \frac{5}{2}$ ; $3 tan^{2} 3 0^{\circ} = 3 \times \frac{1}{3} = 1$ ; $2 sec^{2} 3 0^{\circ} = 2 \times \frac{4}{3} = \frac{8}{3}$
Value $= \frac{\frac{5}{2} - 1}{\frac{8}{3}} = \frac{3}{2} \times \frac{3}{8} = \frac{9}{16}$

Q23 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Basic 430/5/2

For A = $6 0^{\circ}$ and B = $3 0^{\circ}$ , verify that $tan (A - B) = \frac{t a n A - t a n B}{1 + t a n A t a n B}$ .

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Answer: Verified: both sides equal

\frac{1}{3}

LHS $= tan (6 0^{\circ} - 3 0^{\circ}) = tan 3 0^{\circ} = \frac{1}{3}$
RHS $= \frac{3 - \frac{1}{3}}{1 + 3 \times \frac{1}{3}} = \frac{\frac{2}{3}}{2} = \frac{1}{3}$
LHS = RHS. Verified.

Q242 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/1/1

If $tan θ = \frac{24}{7}$ , then find the value of $sin θ + cos θ$ .

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Answer:

\frac{31}{25}

Take opposite side 24k and adjacent side 7k.
Hypotenuse $= (24 k)^{2} + (7 k)^{2} = 625 k^{2} = 25 k$ .
$sin θ = \frac{24}{25}$ , $cos θ = \frac{7}{25}$ .
$sin θ + cos θ = \frac{31}{25}$ .

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

Q24 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/1/1

If $cot θ = \frac{7}{8}$ , then find the value of $\frac{( 1 + s i n θ ) ( 1 - s i n θ )}{( 1 + c o s θ ) ( 1 - c o s θ )}$ .

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Answer:

\frac{49}{64}

$\frac{( 1 + s i n θ ) ( 1 - s i n θ )}{( 1 + c o s θ ) ( 1 - c o s θ )} = \frac{1 - s i n ^{2} θ}{1 - c o s ^{2} θ} = \frac{c o s ^{2} θ}{s i n ^{2} θ}$
$= cot^{2} θ = (\frac{7}{8})^{2} = \frac{49}{64}$ .

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

Q242 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/2/1

If $tan θ + \frac{1}{t a n θ} = 2$ , find the value of $tan^{2} θ + \frac{1}{t a n ^{2} θ}$ .

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Answer: 2

Squaring both sides: $tan^{2} θ + 2 + \frac{1}{t a n ^{2} θ} = 4$
$tan^{2} θ + \frac{1}{t a n ^{2} θ} = 2$

Q24 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/2/1

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

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Answer: Proved.

LHS $= \frac{( 1 - s i n θ ) ( 1 - s i n θ )}{( 1 + s i n θ ) ( 1 - s i n θ )} = \frac{( 1 - s i n θ ) ^{2}}{1 - s i n ^{2} θ}$
$= \frac{( 1 - s i n θ ) ^{2}}{c o s ^{2} θ} = \frac{1 - s i n θ}{c o s θ}$ (since $1 - sin θ \geq 0$ , taking $cos θ > 0$ )
$= \frac{1}{c o s θ} - \frac{s i n θ}{c o s θ} = sec θ - tan θ$ = RHS

Q22 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/2/2

Prove that : $1 + \frac{c o t ^{2} α}{1 + cosec α} = cosec α$

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Answer: Proved.

$cot^{2} α = cose c^{2} α - 1 = (cosec α - 1) (cosec α + 1)$
LHS $= 1 + \frac{( cosec α - 1 ) ( cosec α + 1 )}{1 + cosec α} = 1 + cosec α - 1 = cosec α$ = RHS

Q252 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/2/3

If $tan A = \frac{4}{3}$ , find $sin A$ and $cos A$ .

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Answer:

sin A = \frac{4}{5}

cos A = \frac{3}{5}

Take a right triangle with opposite side 4k and adjacent side 3k.
Hypotenuse $= 16 k^{2} + 9 k^{2} = 5 k$
$sin A = \frac{4}{5}$ , $cos A = \frac{3}{5}$

Q25 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/2/3

Express $cos A$ and $tan A$ in terms of $sin A$ .

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Answer:

cos A = 1 - sin^{2} A

tan A = \frac{s i n A}{1 - s i n ^{2} A}

$sin^{2} A + cos^{2} A = 1$ , so $cos A = 1 - sin^{2} A$ (for acute A).
$tan A = \frac{s i n A}{c o s A} = \frac{s i n A}{1 - s i n ^{2} A}$

Q23 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/4/1

Evaluate : $\frac{3 c o s ^{2} 3 0 ^{\circ} - 6 cosec ^{2} 3 0 ^{\circ}}{t a n ^{2} 6 0 ^{\circ}}$

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Answer:

- \frac{29}{4}

$cos 3 0^{\circ} = \frac{3}{2}$ , $cosec 3 0^{\circ} = 2$ , $tan 6 0^{\circ} = 3$
Numerator $= 3 \times \frac{3}{4} - 6 \times 4 = \frac{9}{4} - 24 = - \frac{87}{4}$
Denominator $= 3$
Value $= - \frac{87}{4} \times \frac{1}{3} = - \frac{29}{4}$

Also asked in: 2026 Standard 30/4/2

Q212 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/4/3

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

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Answer: Proved.

$\frac{t a n θ}{1 + t a n ^{2} θ} = \frac{t a n θ}{s e c ^{2} θ} = \frac{s i n θ}{c o s θ} \times cos^{2} θ = sin θ cos θ$
$\frac{c o t θ}{1 + c o t ^{2} θ} = \frac{c o t θ}{cosec ^{2} θ} = \frac{c o s θ}{s i n θ} \times sin^{2} θ = sin θ cos θ$
LHS $= sin θ cos θ + sin θ cos θ = 2 sin θ cos θ$ = RHS

Q21 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/4/3

Evaluate : $\frac{1 - 2 t a n ^{2} 3 0 ^{\circ} - s e c ^{2} 4 5 ^{\circ}}{s i n ^{2} 6 0 ^{\circ}}$

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Answer:

- \frac{20}{9}

$tan 3 0^{\circ} = \frac{1}{3}$ , $sec 4 5^{\circ} = 2$ , $sin 6 0^{\circ} = \frac{3}{2}$
Numerator $= 1 - \frac{2}{3} - 2 = - \frac{5}{3}$
Denominator $= \frac{3}{4}$
Value $= - \frac{5}{3} \times \frac{4}{3} = - \frac{20}{9}$

Q242 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/5/1

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

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Answer:

\frac{3}{12}

$sin^{3} 6 0^{\circ} = (\frac{3}{2})^{3} = \frac{3 3}{8}$ , $tan 3 0^{\circ} = \frac{1}{3} = \frac{3}{3}$ , $cos^{2} 4 5^{\circ} = \frac{1}{2}$
Numerator $= \frac{3 3}{8} - \frac{3}{3} = \frac{9 3 - 8 3}{24} = \frac{3}{24}$
Value $= \frac{3}{24} \div \frac{1}{2} = \frac{3}{12}$

Also asked in: 2026 Standard 30/5/3

Q24 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/5/1

For acute angles A and B and A + 2B and 2A + B are acute if $tan (A + 2 B) = 3$ and $sin (2 A + B) = \frac{1}{2}$ , then find the measures of angles A and B.

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Answer:

A = 1 0^{\circ}

B = 2 5^{\circ}

$tan (A + 2 B) = 3 = tan 6 0^{\circ}$ , so $A + 2 B = 6 0^{\circ}$ ... (1)
$sin (2 A + B) = \frac{1}{2} = sin 4 5^{\circ}$ , so $2 A + B = 4 5^{\circ}$ ... (2)
Adding: $3 A + 3 B = 10 5^{\circ}$ , so $A + B = 3 5^{\circ}$ ... (3)
(1) − (3): $B = 2 5^{\circ}$ ; then $A = 1 0^{\circ}$ .

Also asked in: 2026 Standard 30/5/3

Q222 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/5/2

For acute angles A and B, if $sec (2 A - B) = 2$ and $cosec (A + B) = 2$ , then find the values of A and B.

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Answer:

A = 2 5^{\circ}

B = 5^{\circ}

$sec (2 A - B) = 2 = sec 4 5^{\circ}$ , so $2 A - B = 4 5^{\circ}$ ... (1)
$cosec (A + B) = 2 = cosec 3 0^{\circ}$ , so $A + B = 3 0^{\circ}$ ... (2)
Adding: $3 A = 7 5^{\circ}$ , so $A = 2 5^{\circ}$ ; then $B = 3 0^{\circ} - 2 5^{\circ} = 5^{\circ}$ .

Q22 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2026 · Standard 30/5/2

Evaluate : $\frac{2 c o s 3 0 ^{\circ} - c o t ^{3} 6 0 ^{\circ}}{t a n 3 0 ^{\circ}}$

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Answer:

\frac{8}{3}

$cos 3 0^{\circ} = \frac{3}{2}$ , $cot 6 0^{\circ} = \frac{1}{3}$ , $tan 3 0^{\circ} = \frac{1}{3}$
Numerator $= 3 - \frac{1}{3 3}$
Value $= (3 - \frac{1}{3 3}) \times 3 = 3 - \frac{1}{3} = \frac{8}{3}$

Q242 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/1/1

Find the values of A and B $(0 \leq A < 9 0^{\circ}, 0 \leq B < 9 0^{\circ})$ , if $tan (A + B) = 1$ and $tan (A - B) = \frac{1}{3}$ .

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Answer:

A = 37. 5^{\circ}

B = 7. 5^{\circ}

$tan (A + B) = 1 = tan 4 5^{\circ}$ , so $A + B = 4 5^{\circ}$ .
$tan (A - B) = \frac{1}{3} = tan 3 0^{\circ}$ , so $A - B = 3 0^{\circ}$ .
Adding: $2 A = 7 5^{\circ}$ , so $A = 37. 5^{\circ}$ ; then $B = 7. 5^{\circ}$ .

Also asked in: 2025 Basic 430/1/2, 2025 Basic 430/1/3

Q24 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/1/1

Prove that $tan 4 5^{\circ} = 1$ geometrically.

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Answer: Proved.

Take $△ A B C$ right-angled at B with $∠ A = 4 5^{\circ}$ .
Then $∠ C = 18 0^{\circ} - 9 0^{\circ} - 4 5^{\circ} = 4 5^{\circ} = ∠ A$ .
Sides opposite equal angles are equal, so BC = AB.
$tan 4 5^{\circ} = tan A = \frac{B C}{A B} = 1$ .

Also asked in: 2025 Basic 430/1/2, 2025 Basic 430/1/3

Q242 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/2/1

Evaluate :
$\frac{c o s 4 5 ^{\circ}}{s e c 3 0 ^{\circ} + cosec 3 0 ^{\circ}}$

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Answer:

\frac{3 2 - 6}{8}

$cos 4 5^{\circ} = \frac{1}{2}$ , $sec 3 0^{\circ} = \frac{2}{3}$ , $cosec 3 0^{\circ} = 2$ .
Denominator $= \frac{2}{3} + 2 = \frac{2 + 2 3}{3}$ .
Value $= \frac{1}{2} \times \frac{3}{2 ( 1 + 3 )} = \frac{3}{2 2 ( 3 + 1 )}$ .
Multiply by $\frac{3 - 1}{3 - 1}$ : $\frac{3 ( 3 - 1 )}{2 2 \times 2} = \frac{3 - 3}{4 2}$ .
$= \frac{( 3 - 3 ) 2}{8} = \frac{3 2 - 6}{8}$ .

Q222 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/2/2

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

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Answer:

\frac{6}{2}

$sin 4 5^{\circ} = \frac{1}{2}$ , $sec 3 0^{\circ} = \frac{2}{3}$ , $tan 3 0^{\circ} = \frac{1}{3}$ .
Denominator $= \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$ .
Value $= \frac{1}{2} \times 3 = \frac{3}{2} = \frac{6}{2}$ .

Q252 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/2/3

Evaluate :
$\frac{s i n ^{2} 4 5 ^{\circ}}{cosec ^{2} 3 0 ^{\circ} - t a n ^{2} 4 5 ^{\circ}}$

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Answer:

\frac{1}{6}

$sin^{2} 4 5^{\circ} = \frac{1}{2}$ , $cosec^{2} 3 0^{\circ} = 4$ , $tan^{2} 4 5^{\circ} = 1$ .
Value $= \frac{\frac{1}{2}}{4 - 1} = \frac{1}{6}$ .

Q242 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/3/1

If $sin 3 A = 1$ , then find the value of $cos 2 A - tan^{2} 4 5^{\circ}$ .

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Answer:

- \frac{1}{2}

$sin 3 A = 1 = sin 9 0^{\circ}$ , so $3 A = 9 0^{\circ}$ and $A = 3 0^{\circ}$ .
$cos 2 A - tan^{2} 4 5^{\circ} = cos 6 0^{\circ} - 1 = \frac{1}{2} - 1 = - \frac{1}{2}$ .

Also asked in: 2025 Basic 430/3/2, 2025 Basic 430/3/3

Q24 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/3/1

If $(sec A + tan A) (1 - sin A) = k cos A$ , then find the value of $k$ .

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Answer:

k = 1

$(sec A + tan A) (1 - sin A) = \frac{1 + s i n A}{c o s A} \times (1 - sin A)$ .
$= \frac{1 - s i n ^{2} A}{c o s A} = \frac{c o s ^{2} A}{c o s A} = cos A$ .
So $k cos A = cos A$ , giving $k = 1$ .

Also asked in: 2025 Basic 430/3/2, 2025 Basic 430/3/3

Q222 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/5/1

Evaluate : $2 tan^{2} 4 5^{\circ} + cos^{2} 3 0^{\circ} - sin^{2} 9 0^{\circ}$ .

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Answer:

\frac{7}{4}

$tan 4 5^{\circ} = 1$ , $cos 3 0^{\circ} = \frac{3}{2}$ , $sin 9 0^{\circ} = 1$ .
$2 (1)^{2} + \frac{3}{4} - 1 = 2 + \frac{3}{4} - 1 = \frac{7}{4}$ .

Also asked in: 2025 Basic 430/5/2

Q22 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/5/1

Verify that $cos 2 A = \frac{1 - t a n ^{2} A}{1 + t a n ^{2} A}$ for $A = 3 0^{\circ}$ .

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Answer: Verified (both sides equal

\frac{1}{2}

LHS $= cos 6 0^{\circ} = \frac{1}{2}$ .
$tan^{2} 3 0^{\circ} = \frac{1}{3}$ .
RHS $= \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \frac{\frac{2}{3}}{\frac{4}{3}} = \frac{1}{2}$ .
LHS = RHS, hence verified.

Also asked in: 2025 Basic 430/5/2

Q252 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/5/3

If sec A = $\frac{25}{7}$ , then find the value of cosec A and tan A.

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Answer: cosec A

= \frac{25}{24}

, tan A

= \frac{24}{7}

$sec A = \frac{25}{7} = \frac{hypotenuse}{adjacent side}$ ; take hypotenuse $= 25 k$ , adjacent side $= 7 k$ .
Opposite side $= (25 k)^{2} - (7 k)^{2} = 576 k^{2} = 24 k$ .
$cosec A = \frac{25 k}{24 k} = \frac{25}{24}$ and $tan A = \frac{24 k}{7 k} = \frac{24}{7}$ .

Q25 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/5/3

Verify that sin (A + B) = sin A cos B + cos A sin B for A = $6 0^{\circ}$ and B = $3 0^{\circ}$ .

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Answer: Verified (both sides equal 1).

LHS $= sin (6 0^{\circ} + 3 0^{\circ}) = sin 9 0^{\circ} = 1$ .
RHS $= sin 6 0^{\circ} cos 3 0^{\circ} + cos 6 0^{\circ} sin 3 0^{\circ} = \frac{3}{2} \times \frac{3}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1$ .
LHS = RHS, hence verified.

Q242 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/6/1

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

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Answer:

\frac{6}{5}

$cos 4 5^{\circ} = \frac{1}{2}$ , $tan 3 0^{\circ} = \frac{1}{3}$ , $sin 6 0^{\circ} = \frac{3}{2}$ .
Denominator $= \frac{1}{3} + \frac{3}{2} = \frac{2 + 3}{2 3} = \frac{5}{2 3}$ .
Value $= \frac{1}{2} \times \frac{2 3}{5} = \frac{2 3}{5 2} = \frac{6}{5}$ .

Also asked in: 2025 Basic 430/6/2, 2025 Basic 430/6/3

Q24 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Basic 430/6/1

Verify that $sin 2 A = \frac{2 t a n A}{1 + t a n ^{2} A}$ , for $A = 3 0^{\circ}$ .

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Answer: Verified: both sides equal

\frac{3}{2}

LHS $= sin 6 0^{\circ} = \frac{3}{2}$ .
RHS $= \frac{2 \times \frac{1}{3}}{1 + \frac{1}{3}} = \frac{\frac{2}{3}}{\frac{4}{3}} = \frac{2}{3} \times \frac{3}{4} = \frac{3}{2 3} = \frac{3}{2}$ .
LHS = RHS, hence verified.

Also asked in: 2025 Basic 430/6/2, 2025 Basic 430/6/3

Q212 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/1/1

If $x cos 6 0^{\circ} + y cos 0^{\circ} + sin 3 0^{\circ} - cot 4 5^{\circ} = 5$ , then find the value of $x + 2 y$ .

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Answer:

x + 2 y = 11

$cos 6 0^{\circ} = \frac{1}{2}$ , $cos 0^{\circ} = 1$ , $sin 3 0^{\circ} = \frac{1}{2}$ , $cot 4 5^{\circ} = 1$ .
$\frac{x}{2} + y + \frac{1}{2} - 1 = 5$
$\frac{x}{2} + y = \frac{11}{2}$
Multiply by 2: $x + 2 y = 11$ .

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

Q21 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/1/1

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

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Answer:

2

$tan^{2} 6 0^{\circ} = (3)^{2} = 3$ .
$sin^{2} 6 0^{\circ} + cos^{2} 3 0^{\circ} = \frac{3}{4} + \frac{3}{4} = \frac{3}{2}$ .
Value $= \frac{3}{3/2} = 2$ .

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

Q232 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/2/1

If $tan A = 3$ ; where A is an acute angle, then find the value of $\frac{s i n ^{2} A}{1 + c o s ^{2} A}$ .

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Answer:

\frac{3}{5}

$tan A = 3$ and A is acute, so $A = 6 0^{\circ}$
$sin^{2} A = \frac{3}{4}$ , $cos^{2} A = \frac{1}{4}$
$\frac{s i n ^{2} A}{1 + c o s ^{2} A} = \frac{3/4}{5/4} = \frac{3}{5}$

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

Q212 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/3/1

If $4 k = tan^{2} 6 0^{\circ} - 2 cosec^{2} 3 0^{\circ} - 2 tan^{2} 3 0^{\circ}$ , then find the value of k.

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Answer:

k = - \frac{17}{12}

$tan^{2} 6 0^{\circ} = 3$ , $cosec^{2} 3 0^{\circ} = 4$ , $tan^{2} 3 0^{\circ} = \frac{1}{3}$ .
$4 k = 3 - 2 (4) - 2 (\frac{1}{3}) = 3 - 8 - \frac{2}{3} = - \frac{17}{3}$ .
$k = - \frac{17}{12}$ .

Also asked in: 2025 Standard 30/3/2

Q212 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/3/3

If $tan A + cot A = 6$ , then find the value of $tan^{2} A + cot^{2} A - 4$ .

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Answer: 30

Squaring, $tan^{2} A + cot^{2} A + 2 tan A cot A = 36$ .
$tan A cot A = 1$ , so $tan^{2} A + cot^{2} A = 34$ .
$tan^{2} A + cot^{2} A - 4 = 30$ .

Q242 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/4/1

Find the value of $x$ for which
$(sin A + cosec A)^{2} + (cos A + sec A)^{2} = x + tan^{2} A + cot^{2} A$

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Answer:

x = 7

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

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

Q24 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/4/1

Evaluate the following :
$\frac{3 s i n 3 0 ^{\circ} - 4 s i n ^{3} 3 0 ^{\circ}}{2 s i n ^{2} 5 0 ^{\circ} + 2 c o s ^{2} 5 0 ^{\circ}}$

Show answer & solution

Answer:

\frac{1}{2}

Numerator $= 3 \times \frac{1}{2} - 4 \times \frac{1}{8} = \frac{3}{2} - \frac{1}{2} = 1$
Denominator $= 2 (sin^{2} 5 0^{\circ} + cos^{2} 5 0^{\circ}) = 2$
Value $= \frac{1}{2}$

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

Q232 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/5/1

It is given that $sin (A - B) = sin A cos B - cos A sin B$ . Use it to find the value of $sin 1 5^{\circ}$ .

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Answer:

sin 1 5^{\circ} = \frac{3 - 1}{2 2} = \frac{6 - 2}{4}

$sin 1 5^{\circ} = sin (4 5^{\circ} - 3 0^{\circ}) = sin 4 5^{\circ} cos 3 0^{\circ} - cos 4 5^{\circ} sin 3 0^{\circ}$
$= \frac{1}{2} \cdot \frac{3}{2} - \frac{1}{2} \cdot \frac{1}{2}$
$= \frac{3 - 1}{2 2} = \frac{6 - 2}{4}$

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

Q23 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/5/1

If $sin A = y$ , then express $cos A$ and $tan A$ in terms of $y$ .

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Answer:

cos A = 1 - y^{2}

tan A = \frac{y}{1 - y ^{2}}

$cos^{2} A = 1 - sin^{2} A = 1 - y^{2} \Rightarrow cos A = 1 - y^{2}$ (A acute)
$tan A = \frac{s i n A}{c o s A} = \frac{y}{1 - y ^{2}}$

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

Q222 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/6/1

If $a sec θ + b tan θ = m$ and $b sec θ + a tan θ = n$ , prove that $a^{2} + n^{2} = b^{2} + m^{2}$

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Answer: Proved.

$m^{2} = a^{2} sec^{2} θ + 2 ab sec θ tan θ + b^{2} tan^{2} θ$
$n^{2} = b^{2} sec^{2} θ + 2 ab sec θ tan θ + a^{2} tan^{2} θ$
$m^{2} - n^{2} = a^{2} (sec^{2} θ - tan^{2} θ) - b^{2} (sec^{2} θ - tan^{2} θ) = a^{2} - b^{2}$ (since $sec^{2} θ - tan^{2} θ = 1$ )
Hence $a^{2} + n^{2} = b^{2} + m^{2}$ .

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

Q22 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2025 · Standard 30/6/1

Use the identity : $sin^{2} A + cos^{2} A = 1$ to prove that $tan^{2} A + 1 = sec^{2} A$ . Hence, find the value of $tan A$ , when $sec A = \frac{5}{3}$ , where A is an acute angle.

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Answer: Proved;

tan A = \frac{4}{3}

Divide $sin^{2} A + cos^{2} A = 1$ by $cos^{2} A$ ( $cos A \neq = 0$ ): $\frac{s i n ^{2} A}{c o s ^{2} A} + 1 = \frac{1}{c o s ^{2} A}$
So $tan^{2} A + 1 = sec^{2} A$ .
$tan^{2} A = sec^{2} A - 1 = \frac{25}{9} - 1 = \frac{16}{9}$
A is acute, so $tan A = \frac{4}{3}$ .

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

Q232 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Basic 430/1/2

In a $△ A B C$ , $∠ A = 9 0^{\circ}$ . If $tan C = 3$ , then find the value of $sin B + cos C - cos^{2} B$ .

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Answer:

\frac{1}{4}

$tan C = 3$ , so $C = 6 0^{\circ}$ and $B = 18 0^{\circ} - 9 0^{\circ} - 6 0^{\circ} = 3 0^{\circ}$ .
$sin 3 0^{\circ} + cos 6 0^{\circ} - cos^{2} 3 0^{\circ} = \frac{1}{2} + \frac{1}{2} - \frac{3}{4}$ .
$= \frac{1}{4}$ .

Q242 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Basic 430/1/3

Evaluate : $sin A cos B + cos A sin B$ ; if $A = 3 0^{\circ}$ and $B = 4 5^{\circ}$ .

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Answer:

\frac{3 + 1}{2 2}

(that is,

\frac{6 + 2}{4}

)

$sin 3 0^{\circ} cos 4 5^{\circ} + cos 3 0^{\circ} sin 4 5^{\circ} = \frac{1}{2} \times \frac{1}{2} + \frac{3}{2} \times \frac{1}{2}$ .
$= \frac{1 + 3}{2 2} = \frac{2 + 6}{4}$ .

Q222 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Basic 430/3/1

Evaluate : $4 sin^{2} 6 0^{\circ} tan^{2} 4 5^{\circ} - 2 sec^{2} 3 0^{\circ} tan^{2} 6 0^{\circ}$

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Answer:

- 5

$sin 6 0^{\circ} = \frac{3}{2}$ , $tan 4 5^{\circ} = 1$ , $sec 3 0^{\circ} = \frac{2}{3}$ , $tan 6 0^{\circ} = 3$ .
$4 \times \frac{3}{4} \times 1 - 2 \times \frac{4}{3} \times 3$
$= 3 - 8 = - 5$ .

Q252 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Basic 430/3/2

Evaluate : $3 sec^{2} 3 0^{\circ} cosec 3 0^{\circ} + tan^{2} 6 0^{\circ} \cdot tan^{2} 4 5^{\circ}$

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Answer: 11

$sec 3 0^{\circ} = \frac{2}{3}$ , $cosec 3 0^{\circ} = 2$ , $tan 6 0^{\circ} = 3$ , $tan 4 5^{\circ} = 1$ .
$3 \times \frac{4}{3} \times 2 + 3 \times 1$
$= 8 + 3 = 11$ .

Q212 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Basic 430/3/3

Evaluate : $5 sin^{2} 4 5^{\circ} - sec 6 0^{\circ} cot^{2} 3 0^{\circ}$

Show answer & solution

Answer:

- \frac{7}{2}

$sin 4 5^{\circ} = \frac{1}{2}$ , $sec 6 0^{\circ} = 2$ , $cot 3 0^{\circ} = 3$ .
$5 \times \frac{1}{2} - 2 \times 3 = \frac{5}{2} - 6 = - \frac{7}{2}$ .

Q252 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Basic 430/4/1

Evaluate :
$sin 3 0^{\circ} cos 6 0^{\circ} + cos 3 0^{\circ} sin 6 0^{\circ} - cot 4 5^{\circ}$

Show answer & solution

Answer: 0

$= \frac{1}{2} \times \frac{1}{2} + \frac{3}{2} \times \frac{3}{2} - 1$
$= \frac{1}{4} + \frac{3}{4} - 1 = 0$ .

Also asked in: 2024 Basic 430/4/2

Q212 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Basic 430/4/3

Evaluate :
$5 sin^{2} 6 0^{\circ} + 3 cos^{2} 3 0^{\circ} - sec^{2} 4 5^{\circ}$

Show answer & solution

Answer: 4

$sin 6 0^{\circ} = cos 3 0^{\circ} = \frac{3}{2}$ and $sec 4 5^{\circ} = 2$ .
$= 5 \times \frac{3}{4} + 3 \times \frac{3}{4} - 2 = \frac{15}{4} + \frac{9}{4} - 2$
$= 6 - 2 = 4$ .

Q212 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Basic 430/5/1

Evaluate : $sin^{2} 3 0^{\circ} + cos^{2} 4 5^{\circ} - cos 0^{\circ} \cdot tan 4 5^{\circ}$

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Answer:

- \frac{1}{4}

$sin 3 0^{\circ} = \frac{1}{2}$ , $cos 4 5^{\circ} = \frac{1}{2}$ , $cos 0^{\circ} = 1$ , $tan 4 5^{\circ} = 1$ .
Value $= \frac{1}{4} + \frac{1}{2} - 1 \times 1$ .
$= \frac{3}{4} - 1 = - \frac{1}{4}$ .

Q222 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Basic 430/5/2

Evaluate : $tan^{2} 3 0^{\circ} - tan^{2} 6 0^{\circ} + cosec^{2} 4 5^{\circ}$

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Answer:

- \frac{2}{3}

$tan 3 0^{\circ} = \frac{1}{3}$ , $tan 6 0^{\circ} = 3$ , $cosec 4 5^{\circ} = 2$ .
Value $= \frac{1}{3} - 3 + 2$ .
$= \frac{1}{3} - 1 = - \frac{2}{3}$ .

Q242 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Basic 430/5/3

If $x = sin 6 0^{\circ} cos 3 0^{\circ} + cos 6 0^{\circ} sin 3 0^{\circ}$ , then find the value of x.

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Answer:

x = 1

$sin 6 0^{\circ} = cos 3 0^{\circ} = \frac{3}{2}$ and $cos 6 0^{\circ} = sin 3 0^{\circ} = \frac{1}{2}$ .
$x = \frac{3}{2} \times \frac{3}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{3}{4} + \frac{1}{4}$ .
$x = 1$ .

Q232 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/1/1

Evaluate : $22 cos 4 5^{\circ} sin 3 0^{\circ} + 23 cos 3 0^{\circ}$

Show answer & solution

Answer: 4

$22 \times \frac{1}{2} \times \frac{1}{2} + 23 \times \frac{3}{2}$
$= 1 + 3$
$= 4$

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

Q23 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/1/1

If $A = 6 0^{\circ}$ and $B = 3 0^{\circ}$ , verify that :
$sin (A + B) = sin A cos B + cos A sin B$

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Answer: Verified (both sides equal 1).

LHS $= sin (6 0^{\circ} + 3 0^{\circ}) = sin 9 0^{\circ} = 1$ .
RHS $= sin 6 0^{\circ} cos 3 0^{\circ} + cos 6 0^{\circ} sin 3 0^{\circ} = \frac{3}{2} \times \frac{3}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1$ .
LHS = RHS, hence verified.

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

Q232 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/2/1

Evaluate : $2 sin^{2} 3 0^{\circ} sec 6 0^{\circ} + tan^{2} 6 0^{\circ}$ .

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Answer: 4

$sin 3 0^{\circ} = \frac{1}{2}$ , $sec 6 0^{\circ} = 2$ , $tan 6 0^{\circ} = 3$
$2 \times \frac{1}{4} \times 2 + 3 = 1 + 3 = 4$

Also asked in: 2024 Standard 30/2/2, 2024 Standard 30/2/3

Q23 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/2/1

If $2 sin (A + B) = 3$ and $cos (A - B) = 1$ , then find the measures of angles A and B. $0 \leq A, B, (A + B) \leq 9 0^{\circ}$ .

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Answer:

A = 3 0^{\circ}

B = 3 0^{\circ}

$sin (A + B) = \frac{3}{2}$ , so $A + B = 6 0^{\circ}$ .
$cos (A - B) = 1$ , so $A - B = 0^{\circ}$ .
Adding: $2 A = 6 0^{\circ}$ , so $A = 3 0^{\circ}$ and $B = 3 0^{\circ}$ .

Also asked in: 2024 Standard 30/2/2, 2024 Standard 30/2/3

Q252 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/3/1

Evaluate : $\frac{c o s 4 5 ^{\circ} + s i n 6 0 ^{\circ}}{s e c 3 0 ^{\circ} + cosec 3 0 ^{\circ}}$ .

Show answer & solution

Answer:

\frac{3 ( 2 + 3 )}{4 ( 3 + 1 )} = \frac{3 2 + 3 3 - 6 - 3}{8}

Numerator: $\frac{1}{2} + \frac{3}{2} = \frac{2 + 3}{2}$ .
Denominator: $\frac{2}{3} + 2 = \frac{2 + 2 3}{3}$ .
Value = $\frac{2 + 3}{2} \times \frac{3}{2 ( 1 + 3 )} = \frac{6 + 3}{4 ( 3 + 1 )}$ .
Multiplying by $\frac{3 - 1}{3 - 1}$ : $\frac{( 6 + 3 ) ( 3 - 1 )}{8} = \frac{3 2 - 6 + 3 3 - 3}{8}$ .

Q212 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/3/2

Evaluate : $\frac{s e c ^{2} 4 5 ^{\circ} - t a n ^{2} 4 5 ^{\circ}}{s i n ^{2} 4 5 ^{\circ}}$

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Answer: 2

$sec 4 5^{\circ} = 2$ , $tan 4 5^{\circ} = 1$ , $sin 4 5^{\circ} = \frac{1}{2}$ .
Numerator = $2 - 1 = 1$ ; denominator = $\frac{1}{2}$ .
Value = $1 \div \frac{1}{2} = 2$ .

Q222 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/3/3

Evaluate : $\frac{5 t a n 6 0 ^{\circ}}{( s i n ^{2} 6 0 ^{\circ} + c o s ^{2} 6 0 ^{\circ} ) t a n 3 0 ^{\circ}}$

Show answer & solution

Answer: 15

$sin^{2} 6 0^{\circ} + cos^{2} 6 0^{\circ} = \frac{3}{4} + \frac{1}{4} = 1$ .
$tan 6 0^{\circ} = 3$ , $tan 3 0^{\circ} = \frac{1}{3}$ .
Value = $\frac{5 3}{1 \times \frac{1}{3}} = 53 \times 3 = 15$ .

Q232 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/4/1

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

Show answer & solution

Answer:

\frac{67}{12}

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

Also asked in: 2024 Standard 30/4/2, 2024 Standard 30/4/3

Q23 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/4/1

If $sin (A - B) = \frac{1}{2}$ , $cos (A + B) = \frac{1}{2}$ ; $0 < A + B \leq 9 0^{\circ}$ , $A > B$ ; find $∠ A$ and $∠ B$ .

Show answer & solution

Answer:

∠ A = 4 5^{\circ}

∠ B = 1 5^{\circ}

$sin (A - B) = \frac{1}{2}$ gives $A - B = 3 0^{\circ}$ .
$cos (A + B) = \frac{1}{2}$ gives $A + B = 6 0^{\circ}$ .
Adding: $2 A = 9 0^{\circ}$ , so $A = 4 5^{\circ}$ ; then $B = 1 5^{\circ}$ .

Also asked in: 2024 Standard 30/4/2, 2024 Standard 30/4/3

Q222 marksVery Short AnswerIntroduction to Trigonometry CBSE 2024 · Standard 30/5/1

Evaluate :
$\frac{2 t a n 3 0 ^{\circ} \cdot s e c 6 0 ^{\circ} \cdot t a n 4 5 ^{\circ}}{1 - s i n ^{2} 6 0 ^{\circ}}$

Show answer & solution

Answer:

\frac{16 3}{3}

(i.e.

\frac{16}{3}

)

Numerator $= 2 \times \frac{1}{3} \times 2 \times 1 = \frac{4}{3}$ .
Denominator $= 1 - (\frac{3}{2})^{2} = 1 - \frac{3}{4} = \frac{1}{4}$ .
Value $= \frac{4}{3} \times 4 = \frac{16}{3} = \frac{16 3}{3}$ .

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

Q222 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/1/1

Evaluate : $tan^{2} 6 0^{\circ} - 2 cosec^{2} 3 0^{\circ} - 2 tan^{2} 3 0^{\circ}$ .

Show answer & solution

Answer:

- \frac{17}{3}

$tan 6 0^{\circ} = 3$ , $cosec 3 0^{\circ} = 2$ , $tan 3 0^{\circ} = \frac{1}{3}$
Expression $= (3)^{2} - 2 (2)^{2} - 2 (\frac{1}{3})^{2}$
$= 3 - 8 - \frac{2}{3} = - \frac{17}{3}$

Q212 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/1/2

Evaluate : $2 (sin^{2} 4 5^{\circ} + cot^{2} 3 0^{\circ}) - 6 (cos^{2} 4 5^{\circ} - tan^{2} 3 0^{\circ})$

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Answer: 6

$sin 4 5^{\circ} = cos 4 5^{\circ} = \frac{1}{2}$ , $cot 3 0^{\circ} = 3$ , $tan 3 0^{\circ} = \frac{1}{3}$
Expression $= 2 (\frac{1}{2} + 3) - 6 (\frac{1}{2} - \frac{1}{3})$
$= 7 - 6 \times \frac{1}{6} = 7 - 1 = 6$

Q212 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/1/3

Evaluate : $\frac{3}{2} tan^{2} 3 0^{\circ} - 2 cos^{2} 9 0^{\circ} - \frac{1}{2} cosec^{2} 3 0^{\circ}$

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Answer:

- \frac{3}{2}

$tan 3 0^{\circ} = \frac{1}{3}$ , $cos 9 0^{\circ} = 0$ , $cosec 3 0^{\circ} = 2$
Expression $= \frac{3}{2} \times \frac{1}{3} - 2 \times 0 - \frac{1}{2} \times 4$
$= \frac{1}{2} - 0 - 2 = - \frac{3}{2}$

Q212 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/4/1

If $sin α = \frac{1}{2}$ , then find the value of $(3 cos α - 4 cos^{3} α)$ .

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Answer: 0

$sin α = \frac{1}{2} \Rightarrow α = 3 0^{\circ}$ , so $cos α = \frac{3}{2}$ .
$3 cos α = \frac{3 3}{2}$ .
$4 cos^{3} α = 4 \times \frac{3 3}{8} = \frac{3 3}{2}$ .
$3 cos α - 4 cos^{3} α = 0$ .

Also asked in: 2023 Basic 430/4/2

Q252 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/4/3

In a right triangle PQR, right angled at Q. If $tan P = 3$ , then evaluate $2 sin P cos P$ .

Diagram for CBSE 2023 Class 10 Maths question 25

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Answer:

\frac{3}{2}

$tan P = 3 = \frac{QR}{P Q}$ ; take $QR = 3 k$ , $P Q = k$ .
$P R = 3 k^{2} + k^{2} = 2 k$ .
$sin P = \frac{3}{2}$ , $cos P = \frac{1}{2}$ .
$2 sin P cos P = 2 \times \frac{3}{2} \times \frac{1}{2} = \frac{3}{2}$ .

Q222 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/5/1

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

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Answer:

\frac{3 3}{2 ( 2 + 3 )} = \frac{6 3 - 9}{2}

Numerator $= \frac{1}{2} + 1 = \frac{3}{2}$ .
Denominator $= \frac{2}{3} + 1 = \frac{2 + 3}{3}$ .
Value $= \frac{3}{2} \times \frac{3}{2 + 3} = \frac{3 3}{2 ( 2 + 3 )}$ .
Rationalising: $\frac{3 3 ( 2 - 3 )}{2} = \frac{6 3 - 9}{2}$ .

Also asked in: 2023 Basic 430/5/2, 2023 Basic 430/5/3

Q22 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/5/1

For A = $3 0^{\circ}$ and B = $6 0^{\circ}$ , verify that :
$sin (A + B) = sin A cos B + cos A sin B$ .

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Answer: Verified: both sides equal 1.

LHS $= sin 9 0^{\circ} = 1$ .
RHS $= sin 3 0^{\circ} cos 6 0^{\circ} + cos 3 0^{\circ} sin 6 0^{\circ} = \frac{1}{2} \cdot \frac{1}{2} + \frac{3}{2} \cdot \frac{3}{2}$ .
$= \frac{1}{4} + \frac{3}{4} = 1$ = LHS. Hence verified.

Also asked in: 2023 Basic 430/5/2, 2023 Basic 430/5/3

Q212 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/6/1

Evaluate : $5 cosec^{2} 4 5^{\circ} - 3 sin^{2} 9 0^{\circ} + 5 cos 0^{\circ}$ .

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Answer: 12

$cosec 4 5^{\circ} = 2$ , $sin 9 0^{\circ} = 1$ , $cos 0^{\circ} = 1$ .
Value $= 5 (2)^{2} - 3 (1)^{2} + 5 (1) = 10 - 3 + 5 = 12$ .

Also asked in: 2023 Basic 430/6/3

Q252 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Basic 430/6/2

Evaluate : $\frac{5 cosec ^{2} 3 0 ^{\circ} - c o s 9 0 ^{\circ}}{4 t a n ^{2} 6 0 ^{\circ}}$

Show answer & solution

Answer:

\frac{5}{3}

$cosec 3 0^{\circ} = 2$ , $cos 9 0^{\circ} = 0$ , $tan 6 0^{\circ} = 3$ .
Value $= \frac{5 \times 4 - 0}{4 \times 3} = \frac{20}{12} = \frac{5}{3}$ .

Q232 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/1/1

If $sin θ + cos θ = 3$ , then find the value of $sin θ \cdot cos θ$ .

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Answer:

sin θ cos θ = 1

Square both sides: $sin^{2} θ + cos^{2} θ + 2 sin θ cos θ = 3$ .
$1 + 2 sin θ cos θ = 3$
$sin θ cos θ = 1$

Also asked in: 2023 Standard 30/1/2

Q23 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/1/1

If $sin α = \frac{1}{2}$ and $cot β = 3$ , then find the value of $cosec α + cosec β$ .

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Answer:

2 + 2

$sin α = \frac{1}{2}$ gives $α = 4 5^{\circ}$ , so $cosec α = 2$ .
$cot β = 3$ gives $β = 3 0^{\circ}$ , so $cosec β = 2$ .
$cosec α + cosec β = 2 + 2$

Also asked in: 2023 Standard 30/1/2

Q212 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/1/3

If $sin θ + sin^{2} θ = 1$ , then prove that $cos^{2} θ + cos^{4} θ = 1$ .

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Answer: Proved.

$sin θ = 1 - sin^{2} θ = cos^{2} θ$ .
$cos^{2} θ + cos^{4} θ = cos^{2} θ + (cos^{2} θ)^{2} = sin θ + sin^{2} θ = 1$ .

Q21 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/1/3

If $tan θ = \frac{1}{7}$ , then show that $\frac{cosec ^{2} θ - s e c ^{2} θ}{cosec ^{2} θ + s e c ^{2} θ} = \frac{3}{4}$ .

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Answer: Proved.

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

Q222 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/2/1

If $4 cot^{2} 4 5^{\circ} - sec^{2} 6 0^{\circ} + sin^{2} 6 0^{\circ} + p = \frac{3}{4}$ , then find the value of p.

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Answer:

p = 0

$cot 4 5^{\circ} = 1$ , $sec 6 0^{\circ} = 2$ , $sin 6 0^{\circ} = \frac{3}{2}$
$4 (1) - 4 + \frac{3}{4} + p = \frac{3}{4}$
$\frac{3}{4} + p = \frac{3}{4}$ , so $p = 0$

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

Q22 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/2/1

If $cos A + cos^{2} A = 1$ , then find the value of $sin^{2} A + sin^{4} A$ .

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Answer: 1

$cos A = 1 - cos^{2} A = sin^{2} A$
So $sin^{4} A = cos^{2} A$ .
$sin^{2} A + sin^{4} A = cos A + cos^{2} A = 1$

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

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

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

Show answer & solution

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

Q25 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/4/1

If A and B are acute angles such that $sin (A - B) = 0$ and $2 cos (A + B) - 1 = 0$ , then find angles A and B.

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Answer:

A = 3 0^{\circ}

B = 3 0^{\circ}

$sin (A - B) = 0 \Rightarrow A - B = 0^{\circ}$ , so $A = B$ .
$2 cos (A + B) - 1 = 0 \Rightarrow cos (A + B) = \frac{1}{2} \Rightarrow A + B = 6 0^{\circ}$
Hence $A = B = 3 0^{\circ}$ .

Q212 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/4/2

Evaluate : $\frac{5}{c o t ^{2} 3 0 ^{\circ}} + \frac{1}{s i n ^{2} 6 0 ^{\circ}} - cot^{2} 4 5^{\circ} + 2 sin^{2} 9 0^{\circ}$

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Answer: 4

$cot 3 0^{\circ} = 3$ , $sin 6 0^{\circ} = \frac{3}{2}$ , $cot 4 5^{\circ} = 1$ , $sin 9 0^{\circ} = 1$
Value $= \frac{5}{3} + \frac{4}{3} - 1 + 2$
$= 3 - 1 + 2 = 4$

Q21 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/4/2

If $θ$ is an acute angle and $sin θ = cos θ$ , find the value of $tan^{2} θ + cot^{2} θ - 2$ .

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Answer: 0

$sin θ = cos θ \Rightarrow tan θ = 1 \Rightarrow θ = 4 5^{\circ}$
$tan^{2} 4 5^{\circ} + cot^{2} 4 5^{\circ} - 2 = 1 + 1 - 2 = 0$

Q222 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/4/3

Evaluate $2 sec^{2} θ + 3 cosec^{2} θ - 2 sin θ cos θ$ if $θ = 4 5^{\circ}$ .

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Answer: 9

At $θ = 4 5^{\circ}$ : $sec^{2} θ = 2$ , $cosec^{2} θ = 2$ , $sin θ cos θ = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}$
Value $= 2 (2) + 3 (2) - 2 (\frac{1}{2}) = 4 + 6 - 1 = 9$

Q22 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/4/3

If $sin θ - cos θ = 0$ , then find the value of $sin^{4} θ + cos^{4} θ$ .

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Answer:

\frac{1}{2}

$sin θ = cos θ \Rightarrow tan θ = 1 \Rightarrow θ = 4 5^{\circ}$ (acute angle)
$sin^{4} θ + cos^{4} θ = (\frac{1}{2})^{4} + (\frac{1}{2})^{4} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

Q252 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/5/1

If $a cos θ + b sin θ = m$ and $a sin θ - b cos θ = n$ , then prove that $a^{2} + b^{2} = m^{2} + n^{2}$ .

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Answer: Proved.

$m^{2} = a^{2} cos^{2} θ + 2 ab sin θ cos θ + b^{2} sin^{2} θ$ .
$n^{2} = a^{2} sin^{2} θ - 2 ab sin θ cos θ + b^{2} cos^{2} θ$ .
Adding: $m^{2} + n^{2} = a^{2} (cos^{2} θ + sin^{2} θ) + b^{2} (sin^{2} θ + cos^{2} θ) = a^{2} + b^{2}$ .
Hence proved.

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

Q25 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry CBSE 2023 · Standard 30/5/1

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

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

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