Question

If $\tan A+\cot A=4$, then ${\tan }^{2}A+{\cot }^{2}A$ is equal to

• A. $10$
• B. $11$
• C. $12$
• D. $14$

Answer 1

The correct option is D $14$

Given, $\tan A+\cot A=4$

On squaring, we get

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

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

$\Rightarrow {\tan }^{2}A+{\cot }^{2}A+2\left(1\right)=16$
$\Rightarrow {\tan }^{2}A+{\cot }^{2}A+2=16$
$\Rightarrow {\tan }^{2}A+{\cot }^{2}A=14$