Question

Prove the following identities (1-16)

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

Answer 1

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