The value of the expression $ \left[\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}\right] $ is
(A) 3
(B) 2
(C) 1
(D) 0

Given:

\( \left[\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}\right] \)

To do:

We have to find the value of the expression \( \left[\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}\right] \).

Solution:

We know that,

$\sin \left(90^{\circ}-\theta\right)=\cos \theta$

$\cos \left(90^{\circ}-\theta\right)=\sin \theta$

$\sin ^{2} \theta+\cos ^{2} \theta=1$

Therefore,

$[\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}]=\frac{\sin ^{2} 22^{\circ}+\sin ^{2}(90^{\circ}-22^{\circ})}{\cos ^{2}(90^{\circ}-68^{\circ})+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin (90^{\circ}-63^{\circ})$

$=\frac{\sin ^{2} 22^{\circ}+\cos ^{2} 22^{\circ}}{\sin ^{2} 68^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} . \cos 63^{\circ}$

$=\frac{1}{1}+(\sin ^{2} 63^{\circ}+\cos ^{2} 63^{\circ})$

$=1+1$

$=2$

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Simply Easy Learning

Updated on: 10-Oct-2022

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