Question

root (sec2θ+cosec2θ)=tanθ+cotθ

Answer 1

Answer:

LHS

$\sqrt{ \sec(2 \alpha ) + \csc(2 \alpha ) } = \sqrt{ \frac{1}{ \cos( 2\alpha ) } + \frac{1}{ \sin( 2\alpha ) } } =$

we know that

$\sin(2 \alpha ) = \frac{2 \tan( \alpha ) }{1 + \tan {}^{2} ( \alpha ) } \: \: and \: \: \: \: \: \: \: \cos(2 \alpha ) = \frac{1 - \tan {}^{2} ( \alpha ) }{1 + \tan {}^{2} ( \alpha ) }$

so from above

$\sqrt{ \frac{1 + \tan {}^{2} ( \alpha ) }{1 - \tan {}^{2} ( \alpha ) } + \frac{1 + \tan {}^{2} ( \alpha ) }{2 \tan( \alpha ) } } =$

$\sqrt{1 + \tan {}^{2} ( \alpha ) } \sqrt{ \frac{2 \tan( \alpha ) + 1 - \tan {}^{2} ( \alpha ) }{2 \tan( \alpha ) (1 - \tan {}^{2} ( \alpha )) } } =$