The value of the expression left[frac{sin ^{2 | vedclass

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Question

(C) Given expression: $\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Given expression: $\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}$Using the identity $\sin(90^{\circ}-	heta) = \cos 	heta$ and $\cos(90^{\circ}-	heta) = \sin 	heta$:Step $1$: Simplify the fraction.$\sin 68^{\circ} = \sin(90^{\circ}-22^{\circ}) = \cos 22^{\circ}$$\cos 22^{\circ} = \cos(90^{\circ}-68^{\circ}) = \sin 68^{\circ}$So,the fraction becomes $\frac{\sin ^{2} 22^{\circ}+\cos ^{2} 22^{\circ}}{\sin ^{2} 68^{\circ}+\cos ^{2} 68^{\circ}} = \frac{1}{1} = 1$.Step $2$: Simplify the remaining terms.$\sin 27^{\circ} = \sin(90^{\circ}-63^{\circ}) = \cos 63^{\circ}$So,$\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ} = \sin ^{2} 63^{\circ}+\cos 63^{\circ} \cdot \cos 63^{\circ} = \sin ^{2} 63^{\circ}+\cos ^{2} 63^{\circ} = 1$.Step $3$: Add the results.$1 + 1 = 2$.

Part $I$	Part $II$
$1.$ $\cos(90^\circ - \theta)$	$a.$ $\sec \theta$
$2.$ $\cot(90^\circ - \theta)$	$b.$ $\sin \theta$
$3.$ $\operatorname{cosec}(90^\circ - \theta)$	$c.$ $1$
	$d.$ $\tan \theta$

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:

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