2023 Trigonometry Questions and Solutions
1. If 2 tan A = 3, then the value of \(\frac{4 \sin A + 3 \cos A}{4 \sin A – 3 \cos A}\) is [CBSE 2023] (1 mark)
(a) \(\frac{7}{\sqrt{13}}\)
(b) \(\frac{1}{\sqrt{13}}\)
(c) 3
(d) does not exist
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Solution:
Given: 2 tan A = 3 ⇒ tan A = \(\frac{3}{2}\)
Divide numerator and denominator by cos A:
\(\frac{4 \sin A + 3 \cos A}{4 \sin A – 3 \cos A} = \frac{4 \tan A + 3}{4 \tan A – 3}\)
Substitute tan A = \(\frac{3}{2}\):
\(= \frac{4 \times \frac{3}{2} + 3}{4 \times \frac{3}{2} – 3} = \frac{6 + 3}{6 – 3} = \frac{9}{3} = 3\)
Answer: (c) 3
2. \(\left[ \frac{5}{8} \sec^2 60^\circ – \tan^2 60^\circ + \cos^2 45^\circ \right]\) is equal to [CBSE 2023] (1 mark)
(a) \(\frac{-5}{3}\)
(b) \(\frac{-1}{2}\)
(c) 0
(d) \(\frac{-1}{4}\)
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Solution:We know:sec 60° = 2 ⇒ sec² 60° = 4
tan 60° = √3 ⇒ tan² 60° = 3
cos 45° = \(\frac{1}{\sqrt{2}}\) ⇒ cos² 45° = \(\frac{1}{2}\)
Now substitute:
\(\frac{5}{8} \times 4 – 3 + \frac{1}{2} = \frac{20}{8} – 3 + \frac{1}{2}\) \(= \frac{5}{2} – 3 + \frac{1}{2} = 3 – 3 = 0\)
Answer: (c) 0
3. (sec²θ – 1)(cosec²θ – 1) is equal to [CBSE 2023] (1 mark)
(a) -1
(b) 1
(c) 0
(d) 2
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Solution:
We know:
sec²θ – 1 = tan²θ
cosec²θ – 1 = cot²θ
So (sec²θ – 1)(cosec²θ – 1) = tan²θ × cot²θ
Since cotθ = \(\frac{1}{\tanθ}\), so tan²θ × cot²θ = 1
Answer: (b) 1
4. Which of the following is true for all values of θ (0° ≤ θ ≤ 90°)? [CBSE 2023] (1 mark)
(a) cos²θ – sin²θ = 1
(b) cosec²θ – sec²θ = 1
(c) sec²θ – tan²θ = 1
(d) cot²θ – tan²θ = 1
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Solution:
This is a standard trigonometric identity:
We know that 1 + tan²θ = sec²θ, so sec²θ – tan²θ = 1
This identity holds true for all values of θ where secθ and tanθ are defined (0° ≤ θ ≤ 90°, θ ≠ 90°).
Answer: (c) sec²θ – tan²θ = 1
5.Evaluate: \(\frac{5}{\cot^2 30^\circ} + \frac{1}{\sin^2 60^\circ} – \cot^2 45^\circ\) \(+ 2\sin^2 90^\circ\) [CBSE 2023] (2 marks)
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Solution:
We know:
\(\cot 30^\circ = \sqrt{3} \Rightarrow \cot^2 30^\circ = 3\)
\(\sin 60^\circ = \frac{\sqrt{3}}{2} \Rightarrow \sin^2 60^\circ = \frac{3}{4}\)
\(\cot 45^\circ = 1 \Rightarrow \cot^2 45^\circ = 1\)
\(\sin 90^\circ = 1 \Rightarrow \sin^2 90^\circ = 1\)
Now substitute:
\(\frac{5}{3} + \frac{1}{\frac{3}{4}} – 1 + 2 \times 1 = \frac{5}{3} + \frac{4}{3} – 1 + 2\)
\(= \frac{9}{3} + 1 = 3 + 1 = 4\)
Answer: 4
6. If θ is an acute angle and sinθ = cosθ, find the value of \(\tan^2 θ + \cot^2 θ – 2\) [CBSE 2023] (2 marks)
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Solution:
Given: sinθ = cosθ
This happens when θ = 45°
Now, tan 45° = 1 and cot 45° = 1
So \(\tan^2 θ + \cot^2 θ – 2 = 1^2 + 1^2 – 2\) \(= 1 + 1 – 2 = 0\)
Answer: 0
7. Take A = 60° and B = 30°. Write the values of cos A + cos B and cos(A + B) [CBSE 2023] (2 marks)
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Solution:
cos A + cos B = cos 60° + cos 30° = \(\frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2}\)
cos(A + B) = cos(60° + 30°) = cos 90° = 0
Answer: cos A + cos B = \(\frac{1 + \sqrt{3}}{2}\), cos(A + B) = 0
8. If sinθ + cosθ = √3, then find the value of sinθ cosθ [CBSE 2023] (2 marks)
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Solution:
Given: sinθ + cosθ = √3
Squaring both sides:
(sinθ + cosθ)² = (√3)²
sin²θ + cos²θ + 2sinθ cosθ = 3
1 + 2sinθ cosθ = 3
2sinθ cosθ = 2
sinθ cosθ = 1
Answer: 1
9. If sinα = \(\frac{1}{\sqrt{2}}\) and cotβ = √3, then find the value of cosecα + cosecβ [CBSE 2023] (2 marks)
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Solution:
sinα = \(\frac{1}{\sqrt{2}}\) ⇒ α = 45° ⇒ cosecα = \(\frac{1}{\sinα} = \sqrt{2}\)
cotβ = √3 ⇒ β = 30° ⇒ sinβ = \(\frac{1}{2}\) ⇒ cosecβ = \(\frac{1}{\sinβ} = 2\)
So cosecα + cosecβ = √2 + 2
Answer: √2 + 2
10. Prove that: \(\frac{\sin A – 2\sin^3 A}{2\cos^3 A – \cos A} = \tan A\) [CBSE 2023] (3 marks)
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Proof:
LHS = \(\frac{\sin A – 2\sin^3 A}{2\cos^3 A – \cos A}\)
= \(\frac{\sin A(1 – 2\sin^2 A)}{\cos A(2\cos^2 A – 1)}\)
Using identity: \(1 – 2\sin^2 A = \cos 2A\) and \(2\cos^2 A – 1 = \cos 2A\)
= \(\frac{\sin A \cdot \cos 2A}{\cos A \cdot \cos 2A}\)
= \(\frac{\sin A}{\cos A}\)
= \(\tan A\) = RHS
Hence proved.
11. Prove that sec A (1 – sin A)(sec A + tan A) = 1 [CBSE 2023] (3 marks)
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Proof:
LHS = sec A (1 – sin A)(sec A + tan A)
= \(\frac{1}{\cos A}(1 – \sin A)\left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)\)
= \(\frac{1}{\cos A}(1 – \sin A)\left(\frac{1 + \sin A}{\cos A}\right)\)
= \(\frac{(1 – \sin A)(1 + \sin A)}{\cos^2 A}\)
= \(\frac{1 – \sin^2 A}{\cos^2 A}\)
= \(\frac{\cos^2 A}{\cos^2 A}\) (since \(1 – \sin^2 A = \cos^2 A\))
= 1 = RHS
Hence proved.
12. Prove that \((\csc A – \sin A)(\sec A – \cos A) \)\(= \frac{1}{\cot A + \tan A}\) [CBSE 2023] (3 marks)
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Proof:
LHS = \((\csc A – \sin A)(\sec A – \cos A)\)
= \(\left(\frac{1}{\sin A} – \sin A\right)\left(\frac{1}{\cos A} – \cos A\right)\)
= \(\left(\frac{1 – \sin^2 A}{\sin A}\right)\left(\frac{1 – \cos^2 A}{\cos A}\right)\)
= \(\left(\frac{\cos^2 A}{\sin A}\right)\left(\frac{\sin^2 A}{\cos A}\right)\)
= \(\sin A \cos A\)
RHS = \(\frac{1}{\cot A + \tan A} = \frac{1}{\frac{\cos A}{\sin A} + \frac{\sin A}{\cos A}}\)
= \(\frac{1}{\frac{\cos^2 A + \sin^2 A}{\sin A \cos A}} = \frac{1}{\frac{1}{\sin A \cos A}} = \sin A \cos A\)
∴ LHS = RHS
Hence proved.