Prove the following identities (1-16)

$\mathrm{LHS}=\frac{\tan ^{3} x}{1+\tan ^{2} x}+\frac{\cot ^{3} x}{1+\cot ^{2} x}$

$=\frac{\tan ^{3} x}{\sec ^{2} x}+\frac{\cot ^{3} x}{\operatorname{cosec}^{2} x}$

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

$=\frac{\sin ^{3} x}{\cos ^{3} x} \times \frac{\cos ^{2} x}{1}+\frac{\cos ^{3} x}{\sin ^{3} x} \times \frac{\sin ^{2} x}{1}$

$=\frac{\sin ^{3} x}{\cos x}+\frac{\cos ^{3} x}{\sin x}$

$=\frac{\sin ^{4} x+\cos ^{4} x}{\sin x \cos x}$

$=\frac{\left(\sin ^{2} x\right)^{2}+\left(\cos ^{2} x\right)^{2}}{\sin x \cos x}$

$=\frac{\left(\sin ^{2} x+\cos ^{2} \mathrm{x}\right)^{2}-2 \sin ^{2} x \cos ^{2} x}{\sin x \cos x}$

$=\frac{1^{2}-2 \sin ^{2} x \cos ^{2} x}{\sin x \cos x}$

$=\frac{1-2 \sin ^{2} x \cos ^{2} x}{\sin x \cos x}$

$=\mathrm{RHS}$

Hence proved.