Question

Prove :

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

Answer 1

Taking $LHS=\frac{1}{\frac{1}{sin\theta }+\frac{cos\theta }{sin\theta }}-\frac{1}{sin\theta }$

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

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

Taking $RHS=\frac{1}{sin\theta }-\frac{1}{\frac{1-cos\theta }{sin\theta }}$

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

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

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

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