Question

If $\alpha$ and $\beta$ are angles in the first quadrant $\tan \alpha =\frac{1}{7},\sin \beta =\frac{1}{\sqrt{10}},$ then using the formula $\sin$ $(A+B)=\sin A\cos B+\cos A\sin B,$ the value of $(\alpha +2\beta )$ is

• A. $0^0$
• B. ${45}^0$
• C. ${60}^0$
• D. ${90}^0$

Answer 1

Answer: (B)

${45}^0$

Given, $\sin \beta =\frac{1}{\sqrt{10}}$
Therefore $\cos \beta =\frac{\sqrt{10-1}}{\sqrt{10}}=\frac{3}{\sqrt{10}}$
Hence, $\tan \beta =\frac{\sin \beta }{cos\beta }=\frac{1}{3}$ ...(i)
$\tan 2\beta =\frac{2\tan \beta }{1-{\tan }^2\beta }$
Substituting the value of $\tan \beta$ from (i), we get
$\tan 2\beta =\frac{3}{4}$ ...(ii)
$\tan \alpha =\frac{1}{7}$ ...(iii)
Now $\tan (\alpha +2\beta )=\frac{\tan \alpha +\tan 2\beta }{1-\tan \alpha +\tan 2\beta }$
Substituting the value of $\tan \alpha$ and $\tan 2\beta$ from (iii) and (ii) and by simplifying, we get
$\tan (\alpha +2\beta )=\frac{4+21}{28-3}$
$=1$
$\tan (\alpha +2\beta )=1$
$\alpha +2\beta ={45}^0$
Hence answer is option B.