Introduction to Trigonometry
CBSE 2026 Board Exam Set 1 (Code 30/1/1)
CBSE_BOARD_PYQ_2026_30_1_1
Grade 10
Question:
[Section B]
Evaluate: $\dfrac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$.
OR
If $\cot \theta = 7/8$, evaluate $\dfrac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}$.
Step-by-Step Solution
Key Concept: Main: Standard values. OR: $(1-\sin^2 \theta)/(1-\cos^2 \theta) = \cos^2 \theta / \sin^2 \theta = \cot^2 \theta = (7/8)^2 = 49/64$.
[Main Question Solution]
Denominator $= 1$. Numerator $= 5(1/4) + 4(4/3) - 1 = 5/4 + 16/3 - 1 = 67/12$. [2.0 Marks]
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[OR Choice Question Solution]
$\dfrac{1 - \sin^2 \theta}{1 - \cos^2 \theta} = \dfrac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta = \left(\dfrac{7}{8}\right)^2 = \dfrac{49}{64}$. [2.0 Marks]
Correct Answer: Main: 67/12 | OR: 49/64
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