If tan A + cot A = 4,

Solution:

(d) 194

We have:

$\tan \mathrm{A}+\cot \mathrm{A}=4$

 

Squaring both the sides:

$(\tan \mathrm{A}+\cot \mathrm{A})^{2}=4^{2}$

$\Rightarrow \tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A}+2(\tan \mathrm{A})(\cot \mathrm{A})=16$

$\Rightarrow \tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A}+2=16$

 

$\Rightarrow \tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A}=14$

Squaring both the sides again:

$\left(\tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A}\right)^{2}=14^{2}$

$\Rightarrow \tan ^{4} \mathrm{~A}+\cot ^{4} \mathrm{~A}+2\left(\tan ^{2} \mathrm{~A}\right)\left(\cot ^{2} \mathrm{~A}\right)=196$

$\Rightarrow \tan ^{4} \mathrm{~A}+\cot ^{4} \mathrm{~A}+2=196$

 

$\Rightarrow \tan ^{4} \mathrm{~A}+\cot ^{4} \mathrm{~A}=194$