2025 Trigonometry Questions and Solutions
1. Which of the following statements is false? (1 Mark)[CBSE 2025]
(A) tan 45° = cot 45°
(B) sin 90° = tan 45°
(C) sin 30° = cos 30°
(D) sin 45° = cos 45°
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Solution:Let’s evaluate each option:(A) tan 45° = 1, cot 45° = 1 → True
(B) sin 90° = 1, tan 45° = 1 → True
(C) sin 30° = \(\frac{1}{2}\), cos 30° = \(\frac{\sqrt{3}}{2}\) → False
(D) sin 45° = \(\frac{1}{\sqrt{2}}\), cos 45° = \(\frac{1}{\sqrt{2}}\) → True
Answer: (C) sin 30° = cos 30°
2. The value of \(\tan^2 A – \frac{1}{\cos^2 A}\) is: (1 Mark)[CBSE 2025]
(A) more than 1
(B) 1
(C) 0
(D) -1
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Solution:
\(\tan^2 A – \frac{1}{\cos^2 A} = \tan^2 A – \sec^2 A\)
We know that \(1 + \tan^2 A = \sec^2 A\), so \(\tan^2 A – \sec^2 A = -1\)
Answer: (D) -1
3. Find the values of A and B (0 ≤ A < 90°, 0 ≤ B < 90°), if (2 Mark)[CBSE 2025]
\(\tan (A + B) = 1\) and \(\tan (A – B) = \frac{1}{\sqrt{3}}\)
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Solution:
We know that:
\(\tan 45^\circ = 1\), so \(A + B = 45^\circ\)
\(\tan 30^\circ = \frac{1}{\sqrt{3}}\), so \(A – B = 30^\circ\)
Adding both equations:
\((A + B) + (A – B) = 45^\circ + 30^\circ\)
\(2A = 75^\circ \Rightarrow A = 37.5^\circ\)
Subtracting the equations:
\((A + B) – (A – B) = 45^\circ – 30^\circ\)
\(2B = 15^\circ \Rightarrow B = 7.5^\circ\)
Answer: A = 37.5°, B = 7.5°
4. Prove that tan 45° = 1 geometrically. (2 Mark)[CBSE 2025]
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Solution:
Consider an isosceles right triangle with both legs equal to 1 unit.
In this triangle, both acute angles are 45°.
By definition, \(\tan \theta = \frac{\text{opposite}}{\text{adjacent}}\)
So \(\tan 45^\circ = \frac{1}{1} = 1\)
Thus, \(\tan 45^\circ = 1\)
5. Prove that: (3 Mark)[CBSE 2025]
\(\frac{1 + \cot^2 A}{1 + \tan^2 A} = \left( \frac{1 – \cot A}{1 – \tan A} \right)^2\)
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Solution:
$$\begin{aligned}
& \frac{1+\cot ^2 A}{1+\tan ^2 A}=\left(\frac{1-\cot A}{1-\tan A}\right)^2 \\
\Rightarrow & \frac{1+\frac{1}{\tan ^2 A}}{1+\tan ^2 A} \quad\left(\frac{1-\frac{1}{\tan A}}{1-\tan A}\right)^2 \\
\Rightarrow & \frac{\frac{\tan ^2 A+1}{\tan ^2 A}}{1+\tan ^2 A} \quad\left(\frac{\frac{\tan A-1}{\tan A}}{1-\tan A}\right)^2 \\
\Rightarrow & \frac{1}{\tan ^2 A} \quad\left(\frac{-1}{\tan A}\right)^2 \\
\Rightarrow & \cot ^2 A \quad \frac{1}{\tan ^2 A}=cot ^2 A \\
&
\end{aligned}$$
6. If θ is an acute angle and 7 + 4 sinθ = 9, then the value of θ is: (1 Mark)[CBSE 2025]
(A) 90°
(B) 30°
(C) 45°
(D) 60°
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Solution:
7 + 4 sinθ = 9
4 sinθ = 2
sinθ = \(\frac{1}{2}\)
Since θ is acute, θ = 30°
Answer: (B) 30°
7. The value of \(\tan^2 \theta – \left( \frac{1}{\cos \theta} \times \sec \theta \right)\) is: (1 Mark)[CBSE 2025]
(A) 1
(B) 0
(C) -1
(D) 2
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Solution:\(\tan^2 \theta – \left( \frac{1}{\cos \theta} \times \sec \theta \right) = \tan^2 \theta – (\sec \theta \times \sec \theta)\)\(= \tan^2 \theta – \sec^2 \theta = -(\sec^2 \theta – \tan^2 \theta) = -1\)
Answer: (C) -1
8. If x cos 60° + y cos 0° + sin 30° cot 45° = 5, then find the value of x + 2y. (2 Mark)[CBSE 2025]
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Solution:
We know:
cos 60° = \(\frac{1}{2}\), cos 0° = 1, sin 30° = \(\frac{1}{2}\), cot 45° = 1
Substituting these values:
\(x \cdot \frac{1}{2} + y \cdot 1 + \frac{1}{2} \cdot 1 = 5\)
\(\frac{x}{2} + y + \frac{1}{2} = 5\)
Multiply both sides by 2:
\(x + 2y + 1 = 10\)
\(x + 2y = 9\)
Answer: x + 2y = 9
9. Evaluate: \(\frac{\tan^2 60^\circ}{\sin^2 60^\circ + \cos^2 30^\circ}\) (2 Mark)[CBSE 2025]
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Solution:
We know:\(\tan 60^\circ = \sqrt{3}\), so \(\tan^2 60^\circ = (\sqrt{3})^2 = 3\)
\(\sin 60^\circ = \frac{\sqrt{3}}{2}\), so \(\sin^2 60^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}\)
\(\cos 30^\circ = \frac{\sqrt{3}}{2}\), so \(\cos^2 30^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}\)
Now, denominator: \(\sin^2 60^\circ + \cos^2 30^\circ = \frac{3}{4} + \frac{3}{4} = \frac{6}{4} = \frac{3}{2}\)
Therefore, \(\frac{\tan^2 60^\circ}{\sin^2 60^\circ + \cos^2 30^\circ} = \frac{3}{\frac{3}{2}} = 3 \times \frac{2}{3} = 2\)
Answer: 2
10. Prove that: \(\frac{\tan \theta}{1 – \cot \theta} + \frac{\cot \theta}{1 – \tan \theta} = 1 + \sec \theta \csc \theta\) (3 Marks)[CBSE 2025]
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Solution:
Left side: \(\frac{\tan \theta}{1 – \cot \theta} + \frac{\cot \theta}{1 – \tan \theta}\)
Express in terms of sin and cos:
\(= \frac{\frac{\sin \theta}{\cos \theta}}{1 – \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 – \frac{\sin \theta}{\cos \theta}}\)
\(= \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta – \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta – \sin \theta}{\cos \theta}}\)
\(= \frac{\sin \theta}{\cos \theta} \times \frac{\sin \theta}{\sin \theta – \cos \theta} + \frac{\cos \theta}{\sin \theta} \times \frac{\cos \theta}{\cos \theta – \sin \theta}\)
\(= \frac{\sin^2 \theta}{\cos \theta (\sin \theta – \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (\cos \theta – \sin \theta)}\)
\(= \frac{\sin^2 \theta}{\cos \theta (\sin \theta – \cos \theta)} – \frac{\cos^2 \theta}{\sin \theta (\sin \theta – \cos \theta)}\)
\(= \frac{1}{\sin \theta – \cos \theta} \left( \frac{\sin^2 \theta}{\cos \theta} – \frac{\cos^2 \theta}{\sin \theta} \right)\)
\(= \frac{1}{\sin \theta – \cos \theta} \left( \frac{\sin^3 \theta – \cos^3 \theta}{\sin \theta \cos \theta} \right)\)
Using identity: \(a^3 – b^3 = (a – b)(a^2 + ab + b^2)\)
\(= \frac{1}{\sin \theta – \cos \theta} \left( \frac{(\sin \theta – \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta \cos \theta} \right)\)
\(= \frac{\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta}{\sin \theta \cos \theta}\)
\(= \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta}\) (since \(\sin^2 \theta + \cos^2 \theta = 1\))
\(= \frac{1}{\sin \theta \cos \theta} + 1 = \csc \theta \sec \theta + 1\)
\(= 1 + \sec \theta \csc \theta\)
Hence proved.
11. Prove that: \(\frac{\sin A + \cos A}{\sin A – \cos A} + \frac{\sin A – \cos A}{\sin A + \cos A} = \frac{2}{2 \sin^2 A – 1}\) (3 Mark)[CBSE 2025]
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Solution:
Left side: \(\frac{\sin A + \cos A}{\sin A – \cos A} + \frac{\sin A – \cos A}{\sin A + \cos A}\)
\(= \frac{(\sin A + \cos A)^2 + (\sin A – \cos A)^2}{(\sin A – \cos A)(\sin A + \cos A)}\)
Numerator: \((\sin A + \cos A)^2 + (\sin A – \cos A)^2\)\(= (\sin^2 A + 2\sin A \cos A + \cos^2 A)\) +\( (\sin^2 A – 2\sin A \cos A + \cos^2 A)\)\(= 2\sin^2 A + 2\cos^2 A\) \(= 2(\sin^2 A + \cos^2 A) = 2(1) = 2\)
Denominator: \((\sin A – \cos A)(\sin A + \cos A) = \sin^2 A – \cos^2 A\)
So left side becomes: \(\frac{2}{\sin^2 A – \cos^2 A}\)
Now, \(\sin^2 A – \cos^2 A = \sin^2 A – (1 – \sin^2 A) \) \(= 2\sin^2 A – 1\)
Therefore, left side = \(\frac{2}{2\sin^2 A – 1}\)
Hence proved.