Let `(sqrt(2)sin alpha)/sqrt(1+cos 2alpha)=1/7` and `sqrt((1-cos2 beta)/2)=1/sqrt(10)` where `alpha,beta in (0,pi/2)`. Then `tan(alpha+2beta)` is equal to

Explanation

To solve the problem, we need to find \( \tan(\alpha + 2\beta) \) given the equations:

1. \( \frac{\sqrt{2} \sin \alpha}{\sqrt{1 + \cos 2\alpha}} = \frac{1}{7} \)
2. \( \sqrt{\frac{1 - \cos 2\beta}{2}} = \frac{1}{\sqrt{10}} \)

### Step 1: Solve for \( \tan \alpha \)

From the first equation, we can manipulate it as follows:

\[ \frac{\sqrt{2} \sin \alpha}{\sqrt{1 + \cos 2\alpha}} = \frac{1}{7} \]

We know that \( \cos 2\alpha = 1 - 2\sin^2 \alpha \), so:

\[ 1 + \cos 2\alpha = 1 + (1 - 2\sin^2 \alpha) = 2 - 2\sin^2 \alpha = 2\cos^2 \alpha \]

Substituting this into our equation gives:

\[ \frac{\sqrt{2} \sin \alpha}{\sqrt{2 \cos^2 \alpha}} = \frac{1}{7} \]

This simplifies to:

\[ \frac{\sqrt{2} \sin \alpha}{\sqrt{2} \cos \alpha} = \frac{1}{7} \]

Thus, we have:

\[ \tan \alpha = \frac{1}{7} \]

### Step 2: Solve for \( \tan 2\beta \)

From the second equation, we have:

\[ \sqrt{\frac{1 - \cos 2\beta}{2}} = \frac{1}{\sqrt{10}} \]

Squaring both sides results in:

\[ \frac{1 - \cos 2\beta}{2} = \frac{1}{10} \]

Multiplying through by 2 gives:

\[ 1 - \cos 2\beta = \frac{2}{10} = \frac{1}{5} \]

Thus:

\[ \cos 2\beta = 1 - \frac{1}{5} = \frac{4}{5} \]

Using the identity \( \sin^2 2\beta + \cos^2 2\beta = 1 \):

\[ \sin^2 2\beta = 1 - \cos^2 2\beta = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \]

Therefore:

\[ \sin 2\beta = \frac{3}{5} \]

Now we can find \( \tan 2\beta \):

\[ \tan 2\beta = \frac{\sin 2\beta}{\cos 2\beta} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

### Step 3: Calculate \( \tan(\alpha + 2\beta) \)

Using the formula for \( \tan(a + b) \):

\[ \tan(\alpha + 2\beta) = \frac{\tan \alpha + \tan 2\beta}{1 - \tan \alpha \tan 2\beta} \]

Substituting the values we found:

\[ \tan(\alpha + 2\beta) = \frac{\frac{1}{7} + \frac{3}{4}}{1 - \left(\frac{1}{7} \cdot \frac{3}{4}\right)} \]

Calculating the numerator:

\[ \frac{1}{7} + \frac{3}{4} = \frac{4 + 21}{28} = \frac{25}{28} \]

Calculating the denominator:

\[ 1 - \frac{3}{28} = \frac{28 - 3}{28} = \frac{25}{28} \]

Thus:

\[ \tan(\alpha + 2\beta) = \frac{\frac{25}{28}}{\frac{25}{28}} = 1 \]

### Final Answer

\[ \tan(\alpha + 2\beta) = 1 \]