Question

solve it by starting from LHS...


Mathematics. ....

Answer 1

LHS:

$Given : \frac{1 - cosA}{1 + cosA}$

On rationalizing, we get

$=> \frac{1 - cosA}{1 + cosA} * \frac{1 - cosA}{1 - cosA}$

$=> \frac{(1 - cosA)^2}{1 - cos^2A}$

$=> \frac{(1 - cosA)^2}{sin^2A}$

$=> (\frac{1}{sinA} - \frac{cosA}{sinA})^2$

$=> \boxed{(cosecA - cotA)^2}$

= RHS

-------------------------------------------------------------------------------------------------------

RHS:

=> (cosecA - cotA)^2

=> (cosecA - cotA)(cosecA - cotA)

= > cosec^2A - 2cotAcosecA + cot^2A

$=> \frac{1}{sin^2A} - \frac{2 * cosA * 1}{sinA * sinA} + cot^2A$

$=> \frac{1}{sin^2A} - \frac{2cosA}{sin^2A} + cot^2A$

$=> \frac{1}{sin^2A} - \frac{2cosA}{sin^2A} + \frac{cos^2A}{sin^2A}$

$=>\frac{1 - 2cosA + cos^2A}{sin^2A}$

$=> \frac{1 - 2cosA + cos^2A}{1 - cos^2A}$

$=> \frac{1 - 2cosA + cos^2A}{(1 - cosA)(1 + cosA)}$

$=> \frac{1 - cosA - cosA + cos^2A}{(1 - cosA)(1 + cosA)}$

$=> \frac{1(1 - cosA) - cosA(1 - cosA)}{(1 - cosA)(1 + cosA)}$

$=> \frac{(1 - cosA)(1 - cosA)}{(1 - cosA)(1 + cosA)}$

$=> \boxed{ \frac{1 - cosA}{1 + cosA}}$

= LHS

Hope it helps!