Question

i make you as brainiest just solve this problem

Answer 1

Answer:

-cos(A)

Step-by-step explanation:

Given,

$\frac{ \cos(\pi - a) \cot( \frac{\pi}{2} + a ) \cos( - a) }{ \tan(\pi + a) \tan( \frac{3\pi}{2} + a ) \sin(2\pi - a) }$

We know that,

$\cos(\pi - \alpha ) = - \cos( \alpha ) \: \: and \: \: \cot( \frac{\pi}{2} + \alpha ) = - \tan( \alpha ) \: \: and \: \: \tan(\pi + \alpha ) = \tan( \alpha ) \: \: and \: \: \tan( \frac{3\pi}{2} + \alpha ) = - \tan( \alpha) \: \: and \: \: \sin(2\pi - \alpha) = - \sin( \alpha )$

On subtituting these, we get,

$\frac{ - \cos(a). - \tan(a). - \cos(a) }{ \tan(a). - \cot(a). - \sin(a) }$

On simplifying, we get,

$= - \cos(a)$