Mathematics · 2022
JEE Main · 26 June 2022, Shift 1 · Q26
If sin^2(10^°) sin (20^°) sin (40^°) sin (50^°) sin (70^°)=α-1/16 sin (10^°), then 16+α^-1 is equal to \ \ \ \.
If $\displaystyle \sin ^2\left(10^{\circ}\right) \sin \left(20^{\circ}\right) \sin \left(40^{\circ}\right) \sin \left(50^{\circ}\right) \sin \left(70^{\circ}\right)=\alpha-\frac{1}{16} \sin \left(10^{\circ}\right)$, then $\displaystyle 16+\alpha^{-1}$ is equal to
\_\_\_\_.
Official answer
80
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.