Question

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

Answer 1

Let us consider $\theta$ As $A$

$L.H.S=\frac{\sin A}{\cot A+\cos ecA}$

$\begin{array}{c}=\frac{\sin A}{\frac{\cos A}{\sin A}+\frac{1}{\sin A}}\\ =\frac{\sin A}{\frac{\cos A+1}{\sin A}}\\ =\frac{{\sin }^{2}A}{\cos A+1}\\ =\frac{{\sin }^{2}A}{\cos A+1}\times \frac{\cos A-1}{\cos A-1}\\ =\frac{{\sin }^{2}A\left(\cos A-1\right)}{{\cos }^{2}A-1}\\ =\frac{{\sin }^{2}A\left(\cos A-1\right)}{-{\sin }^{2}A}\\ =1-\cos A\end{array}$

And,

$R.H.S=2+\frac{\sin A}{\cot A-\cos ecA}$

$\begin{array}{c}=2+\frac{\sin A}{\frac{\cos A}{\sin A}-\frac{1}{\sin A}}\\ =2+\frac{\sin A}{\frac{\cos A-1}{\sin A}}\\ =2+\frac{{\sin }^{2}A}{\cos A-1}\\ =2+\frac{{\sin }^{2}A}{\cos A-1}\times \frac{\cos A+1}{\cos A+1}\\ =2+\frac{{\sin }^{2}A\left(\cos A+1\right)}{{\cos }^{2}A-1}\\ =2-\frac{{\sin }^{2}A\left(\cos A+1\right)}{{\sin }^{2}A}\\ =2-\left(1+\cos A\right)\\ =1-\cos A\end{array}$

Thus, $LHS=RHS$

Hence,

$\frac{\sin A}{\cot A+\cos ecA}=2+\frac{\sin A}{\cot A-\cos ecA}$