Question

$\frac{\sqrt{2}\sin \alpha }{\sqrt{1+\cos 2\alpha }}=\frac{1}{7}$ and $\sqrt{\frac{1-\cos 2\beta }{2}}=\frac{1}{\sqrt{10}},\alpha ,\beta \in \left(0,\frac{\pi }{2}\right)$. Then $\tan (\alpha +2\beta )=$____________

Answer 1

Explanation for the correct option:

Step 1: Evaluating the given expression

Solving the given equations:

$$\begin{gathered} \frac{\sqrt{2}\sin \alpha }{\sqrt{1+\cos 2\alpha }}=\frac{1}{7} \\ \Rightarrow \frac{\sqrt{2}\sin \alpha }{\sqrt{2}\cos \alpha }=\frac{1}{7} \\ \Rightarrow \tan \alpha =\frac{1}{7} \end{gathered}$$

Also,

$$\begin{gathered} \sqrt{\frac{1-\cos 2\beta }{2}}=\frac{1}{\sqrt{10}} \\ \Rightarrow \frac{\sqrt{2}\sin \beta }{\sqrt{2}}=\frac{1}{\sqrt{10}} \\ \Rightarrow \sin \beta =\frac{1}{\sqrt{10}} \\ \Rightarrow \tan \beta =\frac{1}{3}\left[\mathrm{tan\beta }=\frac{\mathrm{sin\beta }}{\mathrm{cos\beta }}=\frac{{\displaystyle \frac{1}{\sqrt{10}}}}{{\displaystyle \frac{3}{\sqrt{10}}}}=\frac{1}{3}\right] \end{gathered}$$

Step 2: Finding the value of $\tan 2\beta$

Now,

$$\begin{gathered} \tan 2\beta =\frac{2\tan \beta }{1-{\tan }^2\beta } \\ =\frac{2\left({\displaystyle \frac{1}{3}}\right)}{1-{\left({\displaystyle \frac{1}{3}}\right)}^2} \\ =\frac{3}{4} \end{gathered}$$

Step 3: Finding the value of $\tan (\alpha +2\beta )$

Therefore,

$$\begin{gathered} \tan (\alpha +2\beta )=\frac{\tan \alpha +\tan 2\beta }{1-\tan \alpha \tan 2\beta } \\ =\frac{\left({\displaystyle \frac{1}{7}}+{\displaystyle \frac{3}{4}}\right)}{1-\left({\displaystyle \frac{1}{7}}\right)\left({\displaystyle \frac{3}{4}}\right)} \\ =1 \end{gathered}$$

Hence, the value of the given expression is $1$.