Question

subject maths tell me answer friends​

Answer 1

Answer:

$\frac{(1 - \sin( \alpha ) )}{(1 + \sin( \alpha ) )} \\ = \frac{(1 - \sin( \alpha ))(1 - \sin( \alpha ) ) }{(1 + \sin( \alpha ))(1 - \sin( \alpha ) ) } \\ = \frac{(1 - \sin( \alpha ) ) {}^{2} }{1 - \sin {}^{2} ( \alpha) } \\ = \frac{(1 + \sin {}^{2} (\alpha ) - 2 \sin( \alpha ) }{ \cos{}^{2} (\alpha) } \\ = \frac{1}{ \cos {}^{2} ( \alpha ) } + \frac{ \sin {}^{2} (\alpha ) }{ \cos{}^{2} (\alpha ) } - \frac{2 \sin( \alpha ) }{ \cos{}^{2} ( \alpha ) } \\ = \sec{}^{2} (\alpha ) + \tan{}^{2} ( \alpha ) - 2 \tan( \alpha ) \sec( \alpha ) \\ = ( \sec( \alpha ) - \tan( \alpha ) ) {}^{2} \: \: (proved) \\ \\ formula \: \: - \\ trigonometry - \frac{1}{ \cos( \alpha ) } = \sec( \alpha ) \\ \\ \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = \tan( \alpha ) \\ \\ \sin {}^{2}( \alpha ) + \cos{}^{2} ( \alpha ) = 1 \\ \cos{}^{2}( \alpha ) = 1 - \sin{}^{2}( \alpha ) \\ \\ \\ algebra \: - x {}^{2} - y {}^{2} = (x + y)(x - y) \\ \\ (x - y) {}^{2} = x {}^{2} + y {}^{2} - 2xy$