Question

pls friends give the correct ans ​

Answer 1

$Given \: \: Question \: \: is \\ \\ \tan( \alpha ) = \frac{3}{4} \\ \\ find \: \: \: \frac{(1 - \cos {}^{2} ( \alpha )) }{(1 + \cos {}^{2} ( \alpha ) )} \\ \\ Answer \: \\ \\ \frac{(1 - \cos {}^{2} ( \alpha ) )}{(1 + \cos {}^{2} ( \alpha ) )} = \frac{ \sin {}^{2} ( \alpha ) }{2 + \sin {}^{2} ( \alpha ) } \\ (becoz \: \: \: \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) = 1) \\ \\ = \frac{ \sin {}^{2} ( \alpha ) }{2 + \sin {}^{2} ( \alpha ) } \\ \\ given \: \: \: \tan( \alpha ) = \frac{3}{4} \\ \tan {}^{2} ( \alpha ) = \frac{9}{16} \: \: \: \: = > \sin {}^{2} ( \alpha ) = \frac{9}{25} \\ \\ = \frac{ \frac{9}{25} }{2 + \frac{9}{25} } \\ \\ = \frac{9}{59} \\ \\ \\ theretofore \: \: \: \\ \frac{(1 - \cos {}^{2} ( \alpha )) }{(1 + \cos {}^{2} ( \alpha )) } = \frac{9}{59}$