Question

If $tan\alpha =\frac{1}{7}andsin\beta =\frac{1}{\sqrt{10}}$ , where $o<\alpha ,\beta <\frac{\pi }{2},then2\beta$ is equal to

• A. $\frac{\pi }{4}-\alpha$
• B. $\frac{3\pi }{4}-\alpha$
• C. $\frac{\pi }{8}-\alpha$
• D. $\frac{3\pi }{4}-\frac{\alpha }{2}$

Answer 1

The correct option is A $\frac{\pi }{4}-\alpha$

$tan\alpha =\frac{1}{7}$

$sin\beta =\frac{1}{\sqrt{10}}$

$cos\beta =\sqrt{1-si{n}^2\beta }=\frac{3}{\sqrt{10}}$

$tan\beta =\frac{1}{3}$

$tan2\beta =\frac{2tan\beta }{1-ta{n}^2\beta }=\frac{3}{4}$

Now,

$tan(2\beta +\alpha )=\frac{tan2\beta +tan\alpha }{1-tan\alpha tan2\beta }$

$=\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{1}{7}\times \frac{3}{4}}$

$=1$

$=tan\frac{\pi }{4}$

$\Rightarrow 2\beta +\alpha =\frac{\pi }{4}$

$\Rightarrow 2\beta =\frac{\pi }{4}-\alpha$