Question

solve this mate.plzz ​

Answer 1

$\star\:\:\:\bf\large\underline\blue{Given\:To\:Prove:-}$

• $\frac{1}{1 + \sin \theta } + \frac{1}{1 - \sin \theta } = \small2 { \sec}^{2} \theta$

$\star\:\:\:\bf\large\underline\blue{Proof:-}$

$\bf\underline\blue{Taking\:LHS:-}$

$\frac{1}{1 + \sin \theta } + \frac{1}{1 - \sin \theta }$

$= \frac{1 - \sin \theta + 1 + \sin \theta }{(1 + \sin \theta)(1 - \sin \theta) }$

$= \frac{2}{1 - { \sin }^{2} \theta}$

Now, we know that,

$\sin^{2} \theta + { \cos}^{2} \theta = 1$

So,

${ \cos}^{2} \theta = 1 - \sin^{2} \theta$

So,

$\frac{2}{1 - { \sin }^{2} \theta}$

$= \frac{2}{ { \cos }^{2} \theta }$

$= 2 \: { \sec}^{2} \theta$

$\bf\underline\blue{Taking\:RHS:-}$

$= 2 { \sec}^{2} \theta$

$\bf\large\blue{Therefore,}$

$\bf\large\blue{LHS=RHS,Proved.}$

Answer 2

Answer:

answer for the given problem is given