Question

Solve the equation :-
$\frac{1-\cos \theta }{1+\cos \theta }=(\text{cosec}\theta -\cot \theta {)}^2$

Answer 1

Multiplying $1-\cos \theta$ to both numerator and denominator

$\frac{1-cos\Theta }{1+cos}\times \frac{1-cos\Theta }{1-cos\Theta }=\frac{(1-cos\Theta {)}^2}{1-co{s}^2\Theta }$

$\therefore si{n}^2\Theta +co{s}^2\Theta =1$

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

$LHS=RHS$