Question

If $\tan x+\cot x=3$, then show that ${\tan }^{4}x+{\cot }^{4}x=47$.

Answer 1

Given,

$\tan x+\cot x=3$

Squaring both sides we get,

${\tan }^{2}x+2+{\cot }^{2}x=9$

or, ${\tan }^{2}x+{\cot }^{2}x=7$

Again squaring both sides we get,

${\tan }^{4}x+2+{\cot }^{4}x=49$

or, ${\tan }^{4}x+{\cot }^{4}x=47$.