Question

Prove that (cot θ - cosec θ)^2 = (1-cos θ/1+cosθ)

Answer 1

hope it helps......:-)

Answer 2

${( \cotθ - \cscθ) }^{2} \\ = {( \frac{ \cosθ }{ \sinθ} - \frac{1}{ \sinθ } )}^{2} \\ = {( \frac{ \cosθ -1 }{ \sinθ } )}^{2} \\ \frac{ (\cosθ -1)( \cosθ -1)}{ { \sin }^{2} θ}$
------(i)


Using Trigonometric identity,
sin^2 θ+ cos^2θ=1


sin^2 θ= 1 -cos^2 θ -----(ii)


By inserting (ii) in (i)

$\frac{( \cosθ - 1)( \cosθ - 1)}{(1 - { \cos}^{2}θ)} \\ \\ = \frac{( \cosθ - 1)( \cosθ - 1) }{(1 - \cosθ)(1 + \cosθ) } \\ = \frac{ - ( \cosθ - 1)( \cosθ - 1)}{( \cosθ - 1)(1 + \cosθ)} \\ \\ = \frac{- ( \cosθ - 1)}{(1 + \cosθ)} \\ = \frac{(1 - \cosθ)}{(1 + \cosθ)} \\ = RHS$



Hence proved.