Question

If A and B are acute angles such that $A+B$ and $A-B$ satisfy the equation ${\tan }^2\theta -4\tan \theta +1=0$, then

• A. $A=\frac{\pi }{4}$
• B. $A=\frac{\pi }{2}$
• C. $B=\frac{\pi }{3}$
• D. $A=\frac{\pi }{6}$

Answer 1

Answer: (A), (D)

The correct options are
A $A=\frac{\pi }{4}$
D $A=\frac{\pi }{6}$

As $\tan (A+B),\tan (A-B)$ roots of ${\tan }^2\theta -4\tan \theta +1=0$
Gives
$\tan (A+B)+\tan (A-B)=4$ ...(1)
$\tan (A+B).\tan (A-B)=1$ ...(2)
$\tan ((A+B)+(A-B))=\frac{\tan (A+B)+\tan (A-B)}{1-\tan (A+B).\tan (A-B)}\tan 2A=\frac{4}{1-1}\Rightarrow 2A=\frac{\pi }{2}\Rightarrow A=\frac{\pi }{4}$
Now from (1)
$\tan (A+B)+\tan (A-B)=4\Rightarrow \frac{\tan A+\tan B}{1-\tan A\tan B}+\frac{\tan A-\tan B}{1+\tan A\tan B}=4\Rightarrow \frac{1+\tan B}{1-\tan B}+\frac{1-\tan B}{1+\tan B}=4\Rightarrow \frac{{(1+\tan B)}^2+{(1-\tan B)}^2}{1-{\tan }^{2}B}=4\Rightarrow \frac{1+{\tan }^{2}B+2\tan B+1+{\tan }^{2}B-2\tan B}{1-{\tan }^{2}B}=4\Rightarrow \frac{1+{\tan }^{2}B}{1-{\tan }^{2}B}=4\Rightarrow \frac{1}{\cos 2B}=2\Rightarrow \cos 2B=\frac{1}{2}\Rightarrow 2B=\frac{\pi }{3}\Rightarrow B=\frac{\pi }{6}$