If `(sqrt(2)cos alpha)/sqrt(1-cos2 alpha)=1/7`and `sqrt((1+cos 2 beta)/2)=1/sqrt10,alpha,beta in (pi/2,pi)`, then `tan (2 alpha - beta)`is equal to ........

Explanation

To solve the problem, we need to find the value of \( \tan(2\alpha - \beta) \) given the conditions:

1. \( \frac{\sqrt{2} \cos \alpha}{\sqrt{1 - \cos 2\alpha}} = \frac{1}{7} \)
2. \( \sqrt{\frac{1 + \cos 2\beta}{2}} = \frac{1}{\sqrt{10}} \)
3. \( \alpha, \beta \in \left(\frac{\pi}{2}, \pi\right) \)

### Step 1: Simplifying the first equation

Starting with the first equation:

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

We know that \( \cos 2\alpha = 2\cos^2 \alpha - 1 \). Thus, we can rewrite \( 1 - \cos 2\alpha \):

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

Substituting this back into the equation gives:

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

This simplifies to:

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

Thus,

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

This implies:

\[ \tan \alpha = 7 \]

### Step 2: Simplifying the second equation

Now, consider the second equation:

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

Squaring both sides gives:

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

Multiplying through by 2:

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

Thus,

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

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

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

Solving for \( \cos^2 \beta \):

\[ 2\cos^2 \beta = \frac{1}{5} \implies \cos^2 \beta = \frac{1}{10} \implies \cos \beta = \frac{1}{\sqrt{10}} \]

### Step 3: Finding \( \tan \beta \)

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

\[ \sin^2 \beta = 1 - \frac{1}{10} = \frac{9}{10} \implies \sin \beta = \frac{3}{\sqrt{10}} \]

Thus,

\[ \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{3}{\sqrt{10}}}{\frac{1}{\sqrt{10}}} = 3 \]

### Step 4: Finding \( \tan 2\alpha \)

Using the double angle formula for tangent:

\[ \tan 2\alpha = \frac{2\tan \alpha}{1 - \tan^2 \alpha} = \frac{2 \cdot 7}{1 - 49} = \frac{14}{-48} = -\frac{7}{24} \]

### Step 5: Finding \( \tan(2\alpha - \beta) \)

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

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

Substituting the values we found:

\[ \tan(2\alpha - \beta) = \frac{-\frac{7}{24} - 3}{1 + \left(-\frac{7}{24}\right)(3)} \]

Calculating the numerator:

\[ -\frac{7}{24} - 3 = -\frac{7}{24} - \frac{72}{24} = -\frac{79}{24} \]

Calculating the denominator:

\[ 1 - \frac{21}{24} = \frac{24 - 21}{24} = \frac{3}{24} = \frac{1}{8} \]

Thus,

\[ \tan(2\alpha - \beta) = \frac{-\frac{79}{24}}{\frac{1}{8}} = -\frac{79}{24} \cdot 8 = -\frac{79}{3} \]

### Final Answer

The value of \( \tan(2\alpha - \beta) \) is:

\[ \boxed{-\frac{79}{3}} \]