Question

Pls ans this question quickly ​

Answer 1

Answer:

Sorry I don't read chapter

Answer 2

Step-by-step explanation:

In the given expression, let θ=α

$1 - \frac{ \sin^{2} ( \alpha ) }{1 + \cos( \alpha ) } + \frac{1 + \cos( \alpha ) }{ \sin( \alpha ) } - \frac{ \sin( \alpha ) }{1 - \cos( \alpha ) } - \frac{1}{ \sec( \alpha ) }$

$\frac{1 + \cos( \alpha ) - \sin ^{2} ( \alpha ) }{1 + \cos( \alpha ) } + \frac{1 + \cos( \alpha ) }{ \sin( \alpha ) } - \frac{ \sin( \alpha ) }{1 - \cos( \alpha ) } - \cos( \alpha )$

$\frac{(1 + \cos( \alpha )) - (1 - \cos^{2} ( \alpha )) }{1 + \cos( \alpha ) } + \frac{(1 - \cos ^{2} ( \alpha ) ) - \sin^{2} ( \alpha ) }{(1 - \cos( \alpha )) \sin( \alpha ) } - \cos( \alpha )$

$\frac{(1 + \cos( \alpha )) - (1 - \cos( \alpha )) (1 + \cos( \alpha )) }{1 + \cos( \alpha ) } + \frac{ \sin^{2} ( \alpha ) - \sin^{2} ( \alpha ) }{ \sin( \alpha ) (1 - \cos( \alpha )) } - \cos( \alpha )$

$= \frac{(1 + \cos( \alpha )) \cos( \alpha ) }{1 + \cos( \alpha ) } - \cos( \alpha )$

$= > \cos( \alpha ) - \cos( \alpha )$

$= 0$