Question

Prove that ${(cosec\theta -cot\theta )}^2=\frac{1-cos\theta }{1+cos\theta }$.

Answer 1

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

$={\left(\frac{1-cos\theta }{sin\theta }\right)}^2$

$=\frac{(1-cos\theta {)}^2}{1-co{s}^2\theta }$

$=\frac{(1-cos\theta )(1-cos\theta )}{(1-cos\theta )(1+cos\theta )}$

$=\frac{1-cos\theta }{1+cos\theta }=RHS$