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JEE MAIN 2026
23-01-2026 S2
Question
Let $\frac{\pi}{2}<\theta<\pi$ and $\cot \theta=-\frac{1}{2 \sqrt{2}}$. Then the value of $$ \sin \left(\frac{15 \theta}{2}\right)(\cos 8 \theta+\sin 8 \theta)+\cos \left(\frac{15 \theta}{2}\right)(\cos 8 \theta-\sin 8 \theta) $$ is equal to
Select the correct option:
A
$\frac{1-\sqrt{2}}{\sqrt{3}}$
B
$-\frac{\sqrt{2}}{\sqrt{3}}$
C
$\frac{\sqrt{2}-1}{\sqrt{3}}$
D
$\frac{\sqrt{2}}{\sqrt{3}}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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Question Tags
JEE Main
Mathematics
Medium
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