Question

Prove that : 1/cosec theta -cot theta - 1/sin theta = 1/sin theta - 1/cosec theta + cot theta

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{1}{\mathrm{cosec}\mathrm{\theta }-\cot \mathrm{\theta }}-\frac{1}{\sin \mathrm{\theta }} \\ =\frac{1}{\mathrm{cosec}\mathrm{\theta }-\cot \mathrm{\theta }}\times \frac{\mathrm{cosec}\mathrm{\theta }+\cot \mathrm{\theta }}{\mathrm{cosec}\mathrm{\theta }+\cot \mathrm{\theta }}-\mathrm{cosec}\mathrm{\theta } \\ =\frac{\mathrm{cosec}\mathrm{\theta }+\cot \mathrm{\theta }}{{\mathrm{cosec}}^2\mathrm{\theta }-{\cot }^2\mathrm{\theta }}-\mathrm{cosec}\mathrm{\theta } \\ =\mathrm{cosec}\mathrm{\theta }+\cot \mathrm{\theta }-\mathrm{cosec}\mathrm{\theta }\left[\mathrm{as},{\mathrm{cosec}}^2\mathrm{\theta }-{\cot }^2\mathrm{\theta }=1\right] \\ =\cot \mathrm{\theta } \\ \mathrm{RHS}=\frac{1}{\sin \mathrm{\theta }}-\frac{1}{{\displaystyle \mathrm{cosec}\mathrm{\theta }+\cot \mathrm{\theta }}} \\ =\mathrm{cosec}\mathrm{\theta }-\frac{1}{\mathrm{cosec}\mathrm{\theta }+\cot \mathrm{\theta }}\times \frac{\mathrm{cosec}\mathrm{\theta }-\cot \mathrm{\theta }}{\mathrm{cosec}\mathrm{\theta }-\cot \mathrm{\theta }} \\ =\mathrm{cosec}\mathrm{\theta }-\frac{\left(\mathrm{cosec}\mathrm{\theta }-\cot \mathrm{\theta }\right)}{{\mathrm{cosec}}^2\mathrm{\theta }-{\cot }^2\mathrm{\theta }} \\ =\mathrm{cosec}\mathrm{\theta }-\mathrm{cosec}\mathrm{\theta }+\cot \mathrm{\theta } \\ =\cot \mathrm{\theta } \\ \mathrm{So},\mathrm{LHS}=\mathrm{RHS} \end{gathered}$$