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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,
$sin\ 25^o\ cos\ 65^o + cos\ 25^o\ sin\ 65^o=sin\ 25^o\ cos\ (90^0 - 25^o) + cos\ 25^o\ sin\ (90^o - 25^o)$
$= sin\ 25^o\ sin\ 25^o + cos\ 25^o\ cos\ 25^o$
$= sin^2\ 25^o + cos^2\ 25^o$
$= 1$