Question

If in a right angle $△ABC$, acute angles $A$ and $B$ satisfy $\tan A+\tan B+{\tan }^{2}A+{\tan }^{2}B=10$, then which of the following is/are correct?

• A. $\tan A=\frac{3+\sqrt{5}}{2}$
• B. $\tan A=\frac{3-\sqrt{5}}{2}$
• C. $\tan A=-2-\sqrt{3}$
• D. $\tan A=-2+\sqrt{3}$

Answer 1

Answer: (A), (B)

The correct options are
A $\tan A=\frac{3+\sqrt{5}}{2}$
B $\tan A=\frac{3-\sqrt{5}}{2}$

As $△ABC$ is right angle, so
$A+B=\frac{\pi }{2}$

Now,
$\tan A+\tan B+{\tan }^{2}A+{\tan }^{2}B=10\Rightarrow \tan A+\tan (\frac{\pi }{2}-A){\tan }^{2}A+{\tan }^2(\frac{\pi }{2}-A)=10\Rightarrow \tan A+\cot A+{\tan }^{2}A+{\cot }^{2}A=10\Rightarrow (\tan A+\cot A{)}^2-2+(\tan A+\cot A)=10\Rightarrow (\tan A+\cot A{)}^2+(\tan A+\cot A)-12=0$
Assuming $(\tan A+\cot A)=t$, so
$\Rightarrow t^2+t-12=0\Rightarrow (t+4)(t-3)=0\Rightarrow t=-4,3$
As $A$ is angle of a triangle, so both $\tan A$ and $\cot A$ are positive, so
$\tan A+\cot A=3\Rightarrow {\tan }^{2}A-3\tan A+1=0\Rightarrow \tan A=\frac{3\pm \sqrt{9-4}}{2}\therefore \tan A=\frac{3\pm \sqrt{5}}{2}$