Question

(cosec thita - cot thita)square = 1- cos thita/ 1+ cos thita

Answer 1

Hope it will help u

Answer 2

*I am using A" instead of Ø*

$\huge\sf{Correct\:Question-}$

( CosecA - CotA )² = $\sf\dfrac{1-CosA}{1+CosA}$

$\huge\sf{Solution-}$

Taking LHS,

$\leadsto$ $({{\sf\dfrac{1}{SinA} - \sf\dfrac{CosA}{SinA} )}}^{2}$

$\leadsto$ $({\sf\dfrac{1-CosA}{SinA}})^{2}$

$\leadsto$ $\sf\dfrac{ ({1 - \cos(A) )}^{2} }{ { \sin(A) }^{2} }$

$\leadsto$ $\sf\dfrac{{(1-CosA)(1-CosA)}}{{1-{Cos}^{2}A}}$

$\leadsto$ $\sf\dfrac{\cancel{(1-CosA)}(1-CosA)}{\cancel{(1-CosA)}(1+CosA)}$

$\leadsto$ $\sf\dfrac{1-CosA}{1+CosA}$

$\leadsto$ RHS

Hence proved!

_______________________

$\large\bold{Identities\:used-}$

★ $\sf\dfrac{1}{SinA}$ = CosecA

★ $\sf\dfrac{CosA}{SinA}$ = CotA

★ (a+b)(a-b) = a² - b²

★ Sin²A + Cos²A = 1