Question

correct and fast plss​

Answer 1

Answer:-

I hope it will help you

Answer 2

 $\frac{1}{1+Sin\theta }$ + $\frac{1}{1-Sin\theta }$ = 2 $sec^{2} \theta$ , L.H.S = R.H.S

Step-by-step explanation:

we have to prove the given equation,  

$\frac{1}{1+Sin\theta }$ + $\frac{1}{1-Sin\theta }$ = 2 $sec^{2} \theta$

Step 1 - Solve L.H.S

by taking L.C.M of these two terms we get

   $\frac{1-Sin\theta+ 1 + Sin\theta }{1 - Sin^{2}\theta }$  

In the denominator we get (${1 - Sin^{2}\theta }$) because  we can solve  

( 1 + Sinθ) (1 - Sinθ ) by using formula (a+b)(a-b) = $a^{2} -b^{2}$

As we know ${1 - Sin^{2}\theta }$ = $Cos^{2} \theta$ that's why we can write

$\frac{2}{Cos^{2} \theta}$ and we can write $\frac{1}{Cos^{2} \theta}$ as $Sec^{2} \theta$ then we get the answer from L.H.S is

2 $Sec^{2} \theta$ .

Step 2 - Comparison of L.H.S with R.H.S

our R.H.S is also 2 $Sec^{2} \theta$  this shows our L.H.S = R.H.S. Hence Proved.