Question

Prove the following identities (1-17)

$\frac{{\tan }^3\mathrm{\theta }}{1+{\tan }^2\mathrm{\theta }}+\frac{{\cot }^3\mathrm{\theta }}{1+{\cot }^2\mathrm{\theta }}=\frac{1-2{\sin }^2\mathrm{\theta }{\cos }^2\mathrm{\theta }}{\sin \mathrm{\theta }\cos \mathrm{\theta }}$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{{\tan }^3\mathrm{\theta }}{1+{\tan }^2\mathrm{\theta }}+\frac{{\cot }^3\mathrm{\theta }}{1+{\cot }^2\mathrm{\theta }} \\ =\frac{{\tan }^3\mathrm{\theta }}{{\sec }^2\mathrm{\theta }}+\frac{{\cot }^3\mathrm{\theta }}{{\mathrm{cosec}}^2\mathrm{\theta }} \\ =\frac{{\displaystyle \frac{{\sin }^3\mathrm{\theta }}{{\cos }^3\mathrm{\theta }}}}{{\displaystyle \frac{1}{{\cos }^2\mathrm{\theta }}}}+\frac{{\displaystyle \frac{{\cos }^3\mathrm{\theta }}{{\sin }^3\mathrm{\theta }}}}{{\displaystyle \frac{1}{{\sin }^2\mathrm{\theta }}}} \\ =\frac{{\sin }^3\mathrm{\theta }}{{\cos }^3\mathrm{\theta }}\times \frac{{\cos }^2\mathrm{\theta }}{1}+\frac{{\cos }^3\mathrm{\theta }}{{\sin }^3\mathrm{\theta }}\times \frac{{\sin }^2\mathrm{\theta }}{1} \\ =\frac{{\sin }^3\mathrm{\theta }}{\cos \mathrm{\theta }}+\frac{{\cos }^3\mathrm{\theta }}{\sin \mathrm{\theta }} \\ =\frac{{\sin }^4\mathrm{\theta }+{\cos }^4\mathrm{\theta }}{\sin \mathrm{\theta }\cos \mathrm{\theta }} \\ =\frac{{\left({\sin }^2\mathrm{\theta }\right)}^2+{\left({\cos }^2\mathrm{\theta }\right)}^2}{\sin \mathrm{\theta }\cos \mathrm{\theta }} \\ =\frac{{\left({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }\right)}^2-2{\sin }^2\mathrm{\theta }{\cos }^2\mathrm{\theta }}{\sin \mathrm{\theta }\cos \mathrm{\theta }} \\ =\frac{1^2-2{\sin }^2\mathrm{\theta }{\cos }^2\mathrm{\theta }}{\sin \mathrm{\theta }\cos \mathrm{\theta }} \\ =\frac{1-2{\sin }^2\mathrm{\theta }{\cos }^2\mathrm{\theta }}{\sin \mathrm{\theta }\cos \mathrm{\theta }} \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$