Question

Prove that $\frac{\sin \theta }{\cot \theta +\csc \theta }=2+\frac{\sin \theta }{\cot \theta -\csc \theta }$

Answer 1

Simplify the RHS of $\frac{\sin \theta }{\cot \theta +\csc \theta }=2+\frac{\sin \theta }{\cot \theta -\csc \theta }$

$2+\frac{\sin \theta }{\cot \theta -\csc \theta }=2+\frac{{\sin }^2\theta }{\cos \theta -1}$

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

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

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

$=1-\cos \theta$

Simplify the LHS of $\frac{\sin \theta }{\cot \theta +\csc \theta }=2+\frac{\sin \theta }{\cot \theta -\csc \theta }$

$\frac{\sin \theta }{\cot \theta +\csc \theta }=\frac{{\sin }^2\theta }{\cos \theta +1}$

$=\frac{1-{\cos }^2\theta }{\cos \theta +1}$

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

$=1-\cos \theta$

Hence $LHS=RHS$.