Question

Prove the identity : [tanθ/ (secθ + 1)] + [cotθ/ (cosecθ + 1)] = cosecθ + secθ - secθ.tanθ

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{\tan \mathrm{\theta }}{\sec \mathrm{\theta }+1}+\frac{\cot \mathrm{\theta }}{\mathrm{cosec}\mathrm{\theta }+1} \\ =\frac{{\displaystyle \frac{\sin \mathrm{\theta }}{\cos \mathrm{\theta }}}}{{\displaystyle \frac{1}{\cos \mathrm{\theta }}}+1}+\frac{{\displaystyle \frac{\cos \mathrm{\theta }}{\sin \mathrm{\theta }}}}{{\displaystyle \frac{1}{\sin \mathrm{\theta }}}+1} \\ =\frac{\sin \mathrm{\theta }}{1+\cos \mathrm{\theta }}+\frac{\cos \mathrm{\theta }}{1+\sin \mathrm{\theta }} \\ =\frac{\sin \mathrm{\theta }}{1+\cos \mathrm{\theta }}\times \frac{1-\cos \mathrm{\theta }}{1-\cos \mathrm{\theta }}+\frac{\cos \mathrm{\theta }}{1+\sin \mathrm{\theta }}\times \frac{1-\sin \mathrm{\theta }}{1-\sin \mathrm{\theta }} \\ =\frac{\sin \mathrm{\theta }\left(1-\cos \mathrm{\theta }\right)}{1-{\cos }^2\mathrm{\theta }}+\frac{\cos \mathrm{\theta }\left(1-\sin \mathrm{\theta }\right)}{1-{\sin }^2\mathrm{\theta }} \\ =\frac{\sin \mathrm{\theta }\left(1-\cos \mathrm{\theta }\right)}{{\sin }^2\mathrm{\theta }}+\frac{\cos \mathrm{\theta }\left(1-\sin \mathrm{\theta }\right)}{{\cos }^2\mathrm{\theta }} \\ =\frac{1-\cos \mathrm{\theta }}{\sin \mathrm{\theta }}+\frac{1-\sin \mathrm{\theta }}{\cos \mathrm{\theta }} \\ =\frac{1}{\sin \mathrm{\theta }}-\frac{\cos \mathrm{\theta }}{\sin \mathrm{\theta }}+\frac{1}{\cos \mathrm{\theta }}-\frac{\sin \mathrm{\theta }}{\cos \mathrm{\theta }} \\ =\frac{1}{\sin \mathrm{\theta }}+\frac{1}{\cos \mathrm{\theta }}-\left(\frac{\cos \mathrm{\theta }}{\sin \mathrm{\theta }}+\frac{\sin \mathrm{\theta }}{\cos \mathrm{\theta }}\right) \\ =\mathrm{cosec}\mathrm{\theta }+\sec \mathrm{\theta }-\left(\frac{{\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }}{\sin \mathrm{\theta }.\cos \mathrm{\theta }}\right) \\ =\mathrm{cosec}\mathrm{\theta }+\sec \mathrm{\theta }-\frac{1}{\sin \mathrm{\theta }.\cos \mathrm{\theta }} \\ =\mathrm{cosec}\mathrm{\theta }+\sec \mathrm{\theta }-\mathrm{cosec}\mathrm{\theta }.\sec \mathrm{\theta } \\ =\mathrm{RHS} \end{gathered}$$