JEE MAIN - Mathematics (2022 - 26th June Morning Shift - No. 17)

If $${\sin ^2}(10^\circ )\sin (20^\circ )\sin (40^\circ )\sin (50^\circ )\sin (70^\circ ) = \alpha - {1 \over {16}}\sin (10^\circ )$$, then $$16 + {\alpha ^{ - 1}}$$ is equal to __________.
Answer
80

Explanation

$$(\sin 10^\circ \,.\,\sin 50^\circ \,.\,\sin 70^\circ )\,.\,(\sin 10^\circ \,.\,\sin 20^\circ \,.\,\sin 40^\circ )$$

$$ = \left( {{1 \over 4}\sin 30^\circ } \right)\,.\,\left[ {{1 \over 2}\sin 10^\circ (\cos 20^\circ - \cos 60^\circ )} \right]$$

$$ = {1 \over {16}}\left[ {\sin 10^\circ \left( {\cos 20^\circ - {1 \over 2}} \right)} \right]$$

$$ = {1 \over {32}}[2\sin 10^\circ \,.\,\cos 20^\circ - \sin 10^\circ ]$$

$$ = {1 \over {32}}[\sin 30^\circ - \sin 10^\circ - \sin 10^\circ ]$$

$$ = {1 \over {64}} - {1 \over {64}}\sin 10^\circ $$

Clearly, $$\alpha = {1 \over {64}}$$

Hence $$16 + {\alpha ^{ - 1}} = 80$$

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