Question

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

Answer 1

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

2.

$\mathfrak{\large{\underline{\underline{</strong><strong>P</strong><strong>r</strong><strong>o</strong><strong>v</strong><strong>e</strong><strong> </strong><strong>that</strong><strong>:-}}}}$

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

$\mathfrak{\large{\underline{\underline{Proof:-}}}}$

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

= $\bold{ {cos}^{2} \alpha [\frac{1}{1 - sin \alpha } + \frac{1}{1 + sin \alpha }] }$

= $\bold{ {cos}^{2} \alpha [\frac{1 + sin \alpha + 1 - sin \alpha }{(1 - sin \alpha )(1 + sin \alpha )}] }$

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

$\boxed{\sf{ 1 - {sin}^{2} \alpha = {cos}^{2} \alpha }}$

= $\bold{ {cos}^{2} \alpha ( \frac{2}{ {cos}^{2} \alpha } )}$

= $\bold{2 }$

★ $\boxed{\sf{\frac{ {cos}^{2} \alpha }{1 - sin \alpha } + \frac{ {cos}^{2} \alpha }{1 + sin \alpha } }= 2}$