Question

Prove that cosec theta minus cot theta whole square equals to 1 minus cos theta upon 1 + cos theta

Answer 1

Answer:

see the above attachment.

Answer 2

ANSWER :-

Question :-

${(\cosec\theta - \cot\theta)}^{2} = \dfrac {1 - \cos\theta}{1+ \cos\theta}$

Let's prove it...!

LHS :-

$\leadsto {{(\cosec\theta - \cot\theta)}^{2}}$

$\leadsto {{[\dfrac{1}{\sin\theta } - \dfrac {\cos\theta}{\sin\theta}]}^{2}}$

$\leadsto {[{\dfrac{1 - \cos\theta }{\sin\theta}]}^{2}}$

$\leadsto {\dfrac {{(1 - \cos\theta)}^{2}}{1 - {\cos}^{2}\theta}}$

(Using identity :- ${\sin}^{2}\theta = 1 - {\cos}^{2}\theta$)

$\leadsto {\dfrac {(1 - \cos\theta)\cancel {(1 - \cos\theta)}}{(1 + \cos\theta)\cancel {(1 - \cos\theta)}}}$

$\leadsto {\dfrac {1 - \cos\theta}{1 + \cos\theta}}$

Hence, proved