# Evaluate:(i) $\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}$(ii) $sin\ 25^o\ cos\ 65^o + cos\ 25^o\ sin\ 65^o$.

To find:

We have to evaluate:

(i) $\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}$

(ii) $sin\ 25^o\ cos\ 65^o + cos\ 25^o\ sin\ 65^o$.

Solution:

(i) $\frac{sin^{2} \ 63^{\circ}\ +\ sin^{2} \ 27^{\circ}\ }{cos^{2} \ 17^{\circ}\ +\ cos^{2} \ 73^{\circ}}$

We know that,

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

Therefore,

$\ sin^{2} \ 63^{\circ}\ =\ sin^{2} \ ( 90^{\circ}\ -\ 27^{\circ}) \ =\ cos^{2} \ 27^{\circ}$

Similarly,

$\ cos^{2} \ 17^{\circ}\ =\ cos^{2} \ ( 90^{\circ}\ -\ 73^{\circ}) \ =\ sin^{2} \ 73^{\circ}$

Hence,

$\ \frac{sin^{2} \ 63^{\circ}\ +\ sin^{2} \ 27^{\circ}\ }{cos^{2} \ 17^{\circ}\ +\ cos^{2} \ 73^{\circ}}=\ \frac{cos^{2} \ 27^{\circ}\ +\ sin^{2} \ 27^{\circ}\ }{sin^{2} \ 73^{\circ}\ +\ cos^{2} \ 73^{\circ}}$

We know that,

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

$\Rightarrow \ \frac{cos^{2} \ 27^{\circ}\ +\ sin^{2} \ 27^{\circ}\ }{sin^{2} \ 73^{\circ}\ +\ cos^{2} \ 73^{\circ}}=\ \frac{1}{1}$

$=1$

Hence proved.

(ii) We know that,

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

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

Therefore,

$sin\ 25^o\ cos\ 65^o + cos\ 25^o\ sin\ 65^o=sin\ 25^o\ cos\ (90^0 - 25^o) + cos\ 25^o\ sin\ (90^o - 25^o)$

$= sin\ 25^o\ sin\ 25^o + cos\ 25^o\ cos\ 25^o$

$= sin^2\ 25^o + cos^2\ 25^o$

$= 1$

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Updated on: 10-Oct-2022

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