Introduction to Trigonometry
CBSE 2026 Board Exam Set 3 (Code 30/1/3)
CBSE_BOARD_PYQ_2026_30_1_3
Grade 10
Question:
[Section B]
If $\tan \theta = \dfrac{24}{7}$, find the value of $\sin \theta + \cos \theta$.
OR
Evaluate: $\dfrac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}$ when $\cot \theta = \dfrac{7}{8}$.
Step-by-Step Solution
Key Concept: Use right triangle / Pythagorean identity.
Main: $\tan \theta = 24/7 \Rightarrow \text{Hypotenuse} = \sqrt{24^2+7^2} = 25$. $\sin \theta = 24/25, \cos \theta = 7/25$. $\sin \theta + \cos \theta = 31/25$. [2.0 Marks]
OR: $\dfrac{1 - \sin^2\theta}{1 - \cos^2\theta} = \dfrac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta = \left(\dfrac{7}{8}\right)^2 = \dfrac{49}{64}$. [2.0 Marks]
Correct Answer: 31/25 OR 49/64
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