Q62 / 160 · AP EAMCET 2024 (18 May Shift 1) · Mathematics

$\sin ^2 76^{\circ}+\sin ^2 16^{\circ}-\sin 76^{\circ} \sin 16^{\circ}=$

  • A
    $0$
  • B
    $\frac{1}{4}$
  • Correct answer
    C
    $\frac{3}{4}$
  • D
    $\frac{4}{3}$

Solution

Since, $\sin ^2 76^{\circ}+\sin ^2 16^{\circ}-\sin 76^{\circ} \sin 16^{\circ}$ $=\frac{1}{2}\left\{2 \sin ^2 76^{\circ}+2 \sin ^2 16^{\circ}-2 \sin 76^{\circ} \sin 16^{\circ}\right\}$ $\begin{aligned} & =\frac{1}{2}\left\{1-\cos 152^{\circ}+1-\cos 32^{\circ}+\cos 92^{\circ}-\cos 60^{\circ}\right\} \\ & =\frac{1}{2}\left\{1-\frac{1}{2}-2 \cos 92^{\circ} \cos 60^{\circ}+\cos 92^{\circ}\right\}\end{aligned}$ $=\frac{1}{2}\left(\frac{3}{2}-\frac{2}{2} \cos 92^{\circ}+\cos 92^{\circ}\right)=\frac{3}{4}$

Hence, the correct answer is option: 'C'.

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