Question

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

Answer 1

Lets take LHS and then equate it to RHS.

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

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

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

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

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

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

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

$=$ RHS

$\Rightarrow$ LHS $=$ RHS

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

Hence, proved.