Question

If $\alpha ,\beta$ are acute angles such that $(\alpha +\beta )$ and $(\alpha -\beta )$ satisfy the equation ${\tan }^2\theta -4\tan \theta +1=0,$ then:

• A. $\alpha =\frac{\pi }{4}$
• B. $\beta =\frac{\pi }{6}$
• C. $\alpha =\frac{\pi }{8}$
• D. $\beta =\frac{\pi }{12}$

Answer 1

Answer: (A), (B)

$\beta =\frac{\pi }{6}$
$\alpha =\frac{\pi }{4}$

$\tan (\alpha +\beta )+\tan (\alpha -\beta )=4$ and $\tan (\alpha +\beta ).\tan (\alpha -\beta )=1$

Hence $\tan (\alpha +\beta )=\cot (\alpha -\beta )=\tan \{\frac{\pi }{2}-(\alpha -\beta \left)\right\}$

$\therefore \alpha =\frac{\pi }{4}$

$\tan (\frac{\pi }{4}+\beta )+\tan (\frac{\pi }{4}-\beta )=4$

$\Rightarrow \frac{1+\tan \beta }{1-tan\beta }+\frac{1-\tan \beta }{1+\tan \beta }=4\Rightarrow \frac{1+{\tan }^2\beta }{1-{\tan }^2\beta }=2\Rightarrow {\tan }^2\beta =\frac{1}{3}$

$\Rightarrow \tan \beta =\frac{1}{\sqrt{3}}\Rightarrow \beta =\frac{\pi }{6}$