Verified Solution • 2025–26 Session moderate

QJEEM22Ma9c72: JEE Main 2022 Mathematics - Class 11 JEEM22Ma9c72 Solution: If , then is equal to __________.

If sin⁡2(10∘)sin⁡(20∘)sin⁡(40∘)sin⁡(50∘)sin⁡(70∘)=α−116sin⁡(10∘){\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+α−116 + {\alpha ^{ - 1}} is equal to __________.

Step-by-Step Verified Solution

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

=(14sin⁡30∘) . [12sin⁡10∘(cos⁡20∘−cos⁡60∘)]= \left( {{1 \over 4}\sin 30^\circ } \right)\,.\,\left[ {{1 \over 2}\sin 10^\circ (\cos 20^\circ - \cos 60^\circ )} \right]

=116[sin⁡10∘(cos⁡20∘−12)]= {1 \over {16}}\left[ {\sin 10^\circ \left( {\cos 20^\circ - {1 \over 2}} \right)} \right]

=132[2sin⁡10∘ . cos⁡20∘−sin⁡10∘]= {1 \over {32}}[2\sin 10^\circ \,.\,\cos 20^\circ - \sin 10^\circ ]

=132[sin⁡30∘−sin⁡10∘−sin⁡10∘]= {1 \over {32}}[\sin 30^\circ - \sin 10^\circ - \sin 10^\circ ]

=164−164sin⁡10∘= {1 \over {64}} - {1 \over {64}}\sin 10^\circ

Clearly, α=164\alpha = {1 \over {64}}

Hence 16+α−1=8016 + {\alpha ^{ - 1}} = 80

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