MCQ
$3 \sin ^2 20^{\circ}-2 \tan ^2 45^{\circ}+3 \sin ^2 70^{\circ}$ is equal to:
  • A
    $0$
  • $1$
  • C
    $2$
  • D
    $-1$

Answer

Correct option: B.
$1$
$3 \sin ^2 20^{\circ}-2 \tan ^2 45^{\circ}+3 \sin ^2 70^{\circ}($ Given $)$
$\Rightarrow 3\left(\sin ^2 20^{\circ}+\sin ^2 70^{\circ}\right)-2 \tan ^2 45^{\circ}$
We know that, $\sin 70^{\circ}=\sin (90-20)^{\circ}=\cos 20^{\circ}$ and $\tan 45^{\circ}=1$
Substituting these values in equation $(1),$
we get, $\Rightarrow 3\left(\sin ^2 20^{\circ}+\cos ^2 20^{\circ}\right)-2$
Also, by an identity $\sin ^2 20^{\circ}+\cos ^2 20^{\circ}=1$
$\Rightarrow 3-2=1$
Hence, the correct option is $(b).$

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