Question

Math lovers plz answer this question ​

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

Show that, $\sf{\:(\csc \theta\:-\:\cot \theta)^2\:=\:\dfrac{(1-\cos \theta)}{(1+\cos \theta)}}$

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

First take L.H.S.

$\mapsto\sf{\:(\csc \theta\:-\:\cot \theta)^2} \\ \\ \small\sf{\orange{\:\:\:\csc \theta\:=\:\dfrac{1}{\sin \theta}\:\:and\:\:\cot \theta\:=\:\dfrac{\cos \theta}{\sin \theta}}} \\ \\ \sf{\:=\:\left(\dfrac{1}{\sin \theta}\:-\:\dfrac{\cos \theta}{\sin \theta}\right)^2} \\ \\ \sf{\:=\:\left(\dfrac{1-\cos \theta}{\sin \theta}\right)^2} \\ \\ \sf{\:=\:\dfrac{(1-\cos \theta)^2}{\sin^2 \theta}}$

Now, Take R.H.S.

$\mapsto\sf{\:\dfrac{(1-\cos \theta)}{(1+\cos \theta)}} \\ \\ \small\sf{\:\:\:\:\:Rationalized\:Denominators} \\ \\ \sf{\:=\:\dfrac{(1-\cos \theta)(1-\cos \theta)}{(1-\cos \theta)(1+\cos \theta)}} \\ \\ \sf{\:=\:\dfrac{(1-\cos \theta)^2}{(1-\cos^2 \theta)}} \\ \\ \small\sf{\orange{\:\:\:\:(1-\cos^2 \theta)\:=\:\sin^2 \theta}} \\ \\ \sf{\:=\:\dfrac{(1-\cos^2 \theta)}{\sin^2 \theta}}$

By, first and second,

$\mapsto\sf{\blue{\:\dfrac{(1-\cos \theta)^2}{\sin^2 \theta}\:=\:\dfrac{(1-\cos \theta)^2}{\sin^2 \theta}}}$

L.H.S. = R.H.S.

That's proved.