Question

prove that:
cosec theta+cot theta/cosec theta-cot theta=1+2cot^​2theta+2cosec theta cottheta

Answer 1

$$\begin{gathered} \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{cosec\theta }+\mathrm{cot\theta }}{\mathrm{cosec\theta }-\mathrm{cot\theta }} \\ =\frac{\mathrm{cosec\theta }+\mathrm{cot\theta }}{\mathrm{cosec\theta }-\mathrm{cot\theta }}\times \frac{\mathrm{cosec\theta }+\mathrm{cot\theta }}{\mathrm{cosec\theta }+\mathrm{cot\theta }} \\ =\frac{{\left(\mathrm{cosec\theta }+\mathrm{cot\theta }\right)}^2}{{\mathrm{cosec}}^2\mathrm{\theta }-{\cot }^2\mathrm{\theta }} \\ ={\left(\mathrm{cosec\theta }+\mathrm{cot\theta }\right)}^2\left[{\mathrm{cosec}}^2\mathrm{\theta }-{\cot }^2\mathrm{\theta }=1\right] \\ ={\mathrm{cosec}}^2\mathrm{\theta }+{\cot }^2\mathrm{\theta }+2\mathrm{cosec\theta }\mathrm{cot\theta } \\ =\left(1+{\cot }^2\mathrm{\theta }\right)++{\cot }^2\mathrm{\theta }+2\mathrm{cosec\theta }\mathrm{cot\theta } \\ =1+2{\cot }^2\mathrm{\theta }+2\mathrm{cosec\theta }\mathrm{cot\theta } \\ =\mathrm{R}.\mathrm{H}.\mathrm{S}. \\ \mathrm{Hence}\mathrm{Proved}. \end{gathered}$$