Question

1/1+cos A +1/1-cos A=2cosec^2A

Answer 1

1/(1+cosA)+1/(1-cosA)=2/(1-cos^2A)=2/sin^2A=2cosec^2A

Answer 2

Question:

$\dfrac{1}{1+cosA} + \dfrac{1}{1-cosA} = 2{cosec}^{2}A$

Solution:

LHS = $\dfrac{1}{1+cosA} + \dfrac{1}{1-cosA}$

RHS = $2{cosec}^{2}A$

LHS = $\dfrac{1}{1+cosA} + \dfrac{1}{1-cosA}$

$= \dfrac{1- cosA + 1 + cosA}{(1+cosA)(1-cosA)}$

$= \dfrac{2}{1 - {cos}^{2}A}$

$= \dfrac{2}{{sin}^{2}A}$

$= 2{cosec}^{2}A$ = RHS

Hence, proved.

☆Identities used☆

• $a^{2} - b^{2} = (a+b)(a-b)$

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

• $cosec\theta = \dfrac{1}{sin\theta}$