Question

prove that :cosec sq. theta /cosec theta -1-cosec sq.theta /cosec theta +1=2sec sq.theta

Answer 1

$$\begin{gathered} \mathrm{To}\mathrm{Prove}:- \\ \frac{{\mathrm{cosec}}^2\mathrm{\theta }}{\mathrm{cosec}\mathrm{\theta }-1}-\frac{{\mathrm{cosec}}^2\mathrm{\theta }}{\mathrm{cosec}\mathrm{\theta }+1}=2{\sec }^2\mathrm{\theta } \\ \mathrm{LHS} \\ {\mathrm{cosec}}^2\mathrm{\theta }\left(\frac{\mathrm{cosec}\mathrm{\theta }+1-\mathrm{cosec}\mathrm{\theta }+1}{\left(\mathrm{cosec}\mathrm{\theta }-1\right)\left(\mathrm{cosec}\mathrm{\theta }+1\right)}\right) \\ =\frac{2{\mathrm{cosec}}^2\mathrm{\theta }}{{\mathrm{cosec}}^2\mathrm{\theta }-1}\left(\mathrm{Since}\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)={\mathrm{a}}^2-{\mathrm{b}}^2\right) \\ =\frac{2{\mathrm{cosec}}^2\mathrm{\theta }}{{\cot }^2\mathrm{\theta }}\left(\mathrm{Since}1+{\cot }^2\mathrm{x}={\mathrm{cosec}}^2\mathrm{x}\right) \\ =\frac{{\displaystyle \frac{2}{{\sin }^2\mathrm{\theta }}}}{{\displaystyle \frac{{\cos }^2\mathrm{\theta }}{{\sin }^2\mathrm{\theta }}}}=\frac{2}{{\cos }^2\mathrm{\theta }}=2{\sec }^2\mathrm{\theta } \\ =\mathrm{RHS} \\ \mathrm{HENCE}\mathrm{PROVED} \end{gathered}$$