Question

If $\tan x+\cot x=3$,then ${\tan }^{4}x+{\cot }^{4}x$= ?

• A. $9$
• B. $47$
• C. $49$
• D. $51$

Answer 1

Answer: (B)

$47$

Given,

$\tan x+\cot x=3\Rightarrow {(\tan x+\cot x)}^2=3^2\Rightarrow {\tan }^{2}x+{\cot }^{2}x+2\tan x\cot x=9\Rightarrow {\tan }^{2}x+{\cot }^{2}x+2=9[\because \cot x=\frac{1}{\tan x}]\Rightarrow {\tan }^{2}x+{\cot }^{2}x=7\Rightarrow {({\tan }^{2}x+{\cot }^{2}x)}^2=7^2\Rightarrow {\tan }^{4}x+{\cot }^{4}x+2{\tan }^{2}x{\tan }^{2}x=49\Rightarrow {\tan }^{4}x+{\cot }^{4}x+2=49\Rightarrow {\tan }^{4}x+{\cot }^{4}x=47$

$\therefore {\tan }^{4}x+{\cot }^{4}x=47$