Question

1/1+cos thita + 1/1-cos thita = 2/ sin^2 thita prove that​

Answer 1

Answer:

The Identity which is used in this question is

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

Firstly will take the LCM and then we can solve this question very easily

as》

We have to prove that

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

Now,

LHS,

$\frac{1}{1 + cos\theta} + \frac{1}{1 - cos\theta} \\ = \frac{1 - \cancel{cos\theta} + 1 \cancel{+ cos \theta}}{({1 + cos\theta})( {1 - cos\theta})} \\ = \frac{2}{1 - {cos}^{2}\theta} \\ = \frac{2}{ {sin}^{2}\theta} = RHS \\ \boxed{\boxed{Hence Proved}}$

Answer 2

1/1+cosQ+1/1-cosQ=2/sin^2

1+cosQ+1-cosQ/1^2-cosQ^2=2/sin^2

2/1-cosQ^2=2/sin^2

we know that 1-cosQ^2=sinQ^2

put value..... !!

2/sinQ^2=2/sinQ^2

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