Question

Solve for $\theta$:-
$\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}=\cos ec\theta -\cot \theta$

Answer 1

LHS

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

$\because {\sin }^2\theta =1-{\cos }^2\theta \&\sqrt{{(1-\cos \theta )}^2}=|1-\cos \theta |=1-\cos \theta$

$\Rightarrow \frac{1-\cos \theta }{\sqrt{{\sin }^2\theta }}=\frac{1-\cos \theta }{\left|\sin \theta \right|}$

RHS

$\csc \theta -\cot \theta =\frac{1}{\sin \theta }-\frac{\cos \theta }{\sin \theta }=\frac{1-\cos \theta }{\sin \theta }$

Equating LHS to RHS, we get

$\frac{1-\cos \theta }{\left|\sin \theta \right|}=\frac{1-\cos \theta }{\sin \theta }$

$\Rightarrow \left|\sin \theta \right|=\sin \theta$

$\because \sin \theta$ is positive

$\therefore \theta \in (0,\pi )$

Hence $\theta \in (2n\pi ,(2n+1\left)\pi \right)$