If tan A + cot A = 4, then {tan ^4}A + {cot ^ | vedclass

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(D) Given: $	an A + \cot A = 4$Squaring both sides:$(	an A + \cot A)^2 = 4^2$$	an^2 A + \cot^2 A + 2 	an A \cot A = 16$Since $	an A \cot A = 1$,w

Vedclass · Accepted Answer

$194$

Vedclass · Answer

(D) Given: $	an A + \cot A = 4$Squaring both sides:$(	an A + \cot A)^2 = 4^2$$	an^2 A + \cot^2 A + 2 	an A \cot A = 16$Since $	an A \cot A = 1$,we have:$	an^2 A + \cot^2 A + 2(1) = 16$$	an^2 A + \cot^2 A = 14$Now,squaring both sides again:$(	an^2 A + \cot^2 A)^2 = 14^2$$	an^4 A + \cot^4 A + 2 	an^2 A \cot^2 A = 196$$	an^4 A + \cot^4 A + 2(1)^2 = 196$$	an^4 A + \cot^4 A = 196 - 2 = 194$Thus,the correct option is $D$.

If $\tan A + \cot A = 4,$ then ${\tan ^4}A + {\cot ^4}A$ is equal to

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