Question

Prove that :$\frac{{\tan }^{3}A-{\cot }^{3}A}{\tan A-\cot A}={\sec }^{2}A+{\cot }^{2}A.$

Answer 1

$\frac{{\tan }^{3}A-{\cot }^{3}A}{\tan A-\cot A}={\sec }^{2}A+{\cot }^{2}A$

L.H.S

$=\frac{{\tan }^{3}A-{\cot }^{3}A}{\tan A-\cot A}$

Apply $a^3-b^3=(a-b)(a^2+ab+b^2)$

$=\frac{(\tan A-\cot A)({\tan }^{2}A+{\cot }^{2}A+{\tan }^{2}A{\cot }^{2}A)}{(\tan A-\cot A)}$

$={\tan }^{2}A+1+{\cot }^{2}A$

$={\sec }^{2}A+{\cot }^{2}A$ $[\because 1+{\tan }^{2}A=se{c}^{2}A]$

$=$ R.H.S