Question

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

Answer 1

We know that, $a^3-b^3=(a-b)(a^2+ab+b^2)$

Hence, $\frac{(\tan A-\cot A)({\tan }^{2}A+{\cot }^{2}A+\tan A\cot A)}{\tan A-\cot A}=({\tan }^{2}A+{\cot }^{2}A+1)=(1+{\tan }^{2}A)+{\cot }^{2}A={\sec }^{2}A+{\cot }^{2}A$