Question

whole root 1-sin theta/1+sin theta = sec theta - tan theta​

Answer 1

Step-by-step explanation:

Let thetha=A.

$lhs = \sqrt{ \frac{1 - \sin(a) }{1 + \sin(a) } \times \frac{1 - \sin(a) }{1 - \sin(a) } }$

$= > \sqrt{ \frac{ {(1 - \sin(a) )}^{2} }{ { \cos}^{2} (a)} } = > \frac{1 - \sin(a) }{ \cos(a) }$

$= \frac{1}{ \cos(a) } - \frac{ \sin(a) }{ \cos(a) } ( \frac{1}{ \cos(a) } = \sec(a) \\ and \frac{ \sin(a) }{ \cos(a) } = \tan(a))$

$= > \sec(a) - \tan(a) = rhs \:$