Question

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

• A. $110$
• B. $191$
• C. $80$
• D. $194$

Answer 1

The correct option is D $194$

$tanA+cotA=4$

(squaring both sides)

$(tanA+cotA{)}^2=4^2$

$\Rightarrow ta{n}^{2}A+co{t}^{2}A+2\left(tanA\right)\left(cotA\right)=16$

$\Rightarrow ta{n}^{2}A+co{t}^{2}A+2=16$

$\Rightarrow ta{n}^{2}A+co{t}^{2}A=16-2$

$\Rightarrow ta{n}^{2}A+co{t}^{2}A=14$

(on squaring both sides)

$\Rightarrow (ta{n}^{2}A-co{t}^{2}A{)}^2=(14{)}^2$

$ta{n}^{4}A+co{t}^{4}A+2\left(ta{n}^{2}A\right)\left(co{t}^{2}A\right)=196$

$ta{n}^{2}A+co{t}^{4}A=196-2$

$=194$

Option D is correct