Question

Prove that: $se{c}^{6}A-{\tan }^{6}A=1+3{\tan }^{2}A+3{\tan }^{4}A$

Answer 1

${\sec }^{6}A-{\tan }^{6}A=1+3{\tan }^{2}A+3{\tan }^{4}A$

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

$\therefore {\sec }^{6}A-{\tan }^{6}A=({\sec }^{2}A-{\tan }^{2}A)({\left({\sec }^{2}A\right)}^2+{\sec }^{2}A{\tan }^{2}A+{\left({\tan }^{2}A\right)}^2)$

${\sec }^{6}A-{\tan }^{6}A=((1+{\tan }^{2}A)-{\tan }^{2}A)({(1+{\tan }^{2}A)}^2+(1+{\tan }^{2}A){\tan }^{2}A+{\tan }^{4}A)[\because {\sec }^{2}A=1+{\tan }^{2}A]$

${\sec }^{6}A-{\tan }^{6}A=(1+{\tan }^{2}A-{\tan }^{2}A)(1+{\tan }^{4}A+2{\tan }^{2}A+{\tan }^{2}A+{\tan }^{4}A+{\tan }^{4}A)$

${\sec }^{6}A-{\tan }^{6}A=1+3{\tan }^{2}A+3{\tan }^{4}A$

L.H.S. = R.H.S.

Hence proved