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Evaluate$\frac{sin^{2} \ 63\ +\ sin^{2} \ 27\ }{cos^{2} \ 17\ +\ cos^{2} \ 73}$
Given: $\frac{sin^{2} \ 63\ +\ sin^{2} \ 27\ }{cos^{2} \ 17\ +\ cos^{2} \ 73}$
$\frac{sin^{2} \ 63\ +\ sin^{2} \ 27\ }{cos^{2} \ 17\ +\ cos^{2} \ 73}$
To find: We have to evaluate $\frac{sin^{2} \ 63\ +\ sin^{2} \ 27\ }{cos^{2} \ 17\ +\ cos^{2} \ 73}$
Solution:
$\frac{sin^{2} \ 63\ +\ sin^{2} \ 27\ }{cos^{2} \ 17\ +\ cos^{2} \ 73}$
We know that, $cos( 90\ -\ \Theta ) \ =\ sin\ \Theta \ and\ sin( 90\ -\ \Theta ) \ =\ cos\ \Theta $
$Therefore,\ sin^{2} \ 63\ =\ sin^{2} \ ( 90\ -\ 27) \ =\ cos^{2} \ 27$
$Similarly,\ cos^{2} \ 17\ =\ cos^{2} \ ( 90\ -\ 73) \ =\ sin^{2} \ 73$
$Hence,\ \frac{sin^{2} \ 63\ +\ sin^{2} \ 27\ }{cos^{2} \ 17\ +\ cos^{2} \ 73} \ $
$=\ \frac{cos^{2} \ 27\ +\ sin^{2} \ 27\ }{sin^{2} \ 73\ +\ cos^{2} \ 73}$
$We\ know\ that\ cos^{2} \ \Theta \ \ +\ sin^{2} \ \Theta \ \ =\ 1$
$\Longrightarrow \ \frac{cos^{2} \ 27\ +\ sin^{2} \ 27\ }{sin^{2} \ 73\ +\ cos^{2} \ 73}$
$\Longrightarrow \ \frac{1}{1} \ =\ 1$
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