Question

\[ \text{If } \tan x+\cot x=4, \]

\[ \text{then } \tan^4x+\cot^4x= \]

Solution

Given,

\[ \tan x+\cot x=4 \]

Squaring,

\[ \tan^2x+\cot^2x+2\tan x\cot x=16 \]

\[ \tan x\cot x=1 \]

\[ \tan^2x+\cot^2x+2=16 \]

\[ \tan^2x+\cot^2x=14 \]

Again squaring,

\[ (\tan^2x+\cot^2x)^2 = \tan^4x+\cot^4x+2\tan^2x\cot^2x \]

\[ 14^2 = \tan^4x+\cot^4x+2 \]

\[ 196 = \tan^4x+\cot^4x+2 \]

Therefore,

\[ \tan^4x+\cot^4x=194 \]

Answer

\[ \boxed{194} \]

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