If \sin (20+x)^{\circ}=\cos 60^{\circ}, 0^{\circ} \leq(20+x) \leq 90^{\circ}, then find the value of 2 \sin ^{2}(3 x+15)^{\circ}-cosec^{2}(2 x+10)^{\circ}.

SSC CGL 20 Aug 2021 Shift 1 Paper

Question : 56 of 100

Marks: +1, -0

\sin (20+x)^{∘}=\cos 60^{∘}

0^{∘} ≤(20+x) ≤ 90^{∘}

, then find the value of

2 \sin ^{2}(3 x+15)^{∘}-cosec^{2}(2 x+10)^{∘}

Solution:

Given,

\;\sin (20+x)^{∘}=\cos 60^{∘}

⇒ \;\; \sin (20+x)^{∘}=\sin (90^{∘}-60^{∘})

⇒ \;\; 20^{∘}+x=90^{∘}-60^{∘}

⇒ \;\; x=30^{∘}-20^{∘}=10^{∘}

∴ 2 \sin ^{2}(3 x+15)^{∘}-cosec^{2}(2 x+10)^{∘}

=\;2 \sin ^{2}(3 × 10+15)^{∘}

=\;-cosec^{2}(2 × 10+10)^{∘}

\;=2 ×(\;{1}/{√{2}})^{∘}-(2)^{2}

\;=1-4=-3

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