If $$\frac{\sqrt{2}\sin\alpha}{\sqrt{1+\cos 2\alpha}} = \frac{1}{7}$$ and $$\sqrt{\frac{1-\cos 2\beta}{2}} = \frac{1}{\sqrt{10}}$$, $$\alpha, \beta \in \left(0, \frac{\pi}{2}\right)$$, then $$\tan(\alpha + 2\beta)$$ is equal to

We have been given two separate relations and both angles lie in the first quadrant, that is $$0<\alpha<\dfrac{\pi}{2}$$ and $$0<\beta<\dfrac{\pi}{2}$$, so all trigonometric ratios of these angles are positive.

First of all, consider the equation

$$\dfrac{\sqrt{2}\,\sin\alpha}{\sqrt{1+\cos 2\alpha}}=\dfrac{1}{7}.$$

We recall the double-angle identity

$$\cos 2\alpha = 2\cos^2\alpha - 1,$$

which can be rearranged to give

$$1 + \cos 2\alpha = 2\cos^2\alpha.$$

Substituting this into the denominator we obtain

$$\sqrt{1+\cos 2\alpha} = \sqrt{2\cos^2\alpha} = \sqrt{2}\,\cos\alpha,$$

because $$\cos\alpha>0$$ in the given interval and therefore the absolute value is not required.

Now we rewrite the left‐hand side of the original equation:

$$\dfrac{\sqrt{2}\,\sin\alpha}{\sqrt{2}\,\cos\alpha} = \dfrac{\sin\alpha}{\cos\alpha} = \tan\alpha.$$

Thus we deduce

$$\tan\alpha = \dfrac{1}{7}.$$

Next, consider the second relation

$$\sqrt{\dfrac{1-\cos 2\beta}{2}} = \dfrac{1}{\sqrt{10}}.$$

We recall the familiar identity

$$\sin^2\beta = \dfrac{1-\cos 2\beta}{2}.$$

Taking the positive square root (again justified by $$0<\beta<\dfrac{\pi}{2}$$) we get

$$\sin\beta = \dfrac{1}{\sqrt{10}}.$$

Using the Pythagorean relation $$\sin^2\beta + \cos^2\beta = 1,$$ we find

$$\cos\beta = \sqrt{1 - \sin^2\beta} = \sqrt{1 - \dfrac{1}{10}} = \sqrt{\dfrac{9}{10}} = \dfrac{3}{\sqrt{10}}.$$

Hence

$$\tan\beta = \dfrac{\sin\beta}{\cos\beta} = \dfrac{\dfrac{1}{\sqrt{10}}}{\dfrac{3}{\sqrt{10}}} = \dfrac{1}{3}.$$

Our goal is to compute $$\tan(\alpha + 2\beta).$$ We already know $$\tan\alpha$$ and $$\tan\beta,$$ so we next find $$\tan 2\beta.$$ The double-angle formula for tangent is

$$\tan 2\beta = \dfrac{2\tan\beta}{1-\tan^2\beta}.$$

Substituting $$\tan\beta = \dfrac{1}{3}$$ gives

$$\tan 2\beta = \dfrac{2\left(\dfrac{1}{3}\right)}{1-\left(\dfrac{1}{3}\right)^2} = \dfrac{\dfrac{2}{3}}{1-\dfrac{1}{9}} = \dfrac{\dfrac{2}{3}}{\dfrac{8}{9}} = \dfrac{2}{3}\times\dfrac{9}{8} = \dfrac{3}{4}.$$

Now we use the angle‐addition formula for tangent:

$$\tan(\alpha + 2\beta) = \dfrac{\tan\alpha + \tan 2\beta}{1 - \tan\alpha\,\tan 2\beta}.$$

We substitute $$\tan\alpha = \dfrac{1}{7}$$ and $$\tan 2\beta = \dfrac{3}{4}$$:

First the numerator:

$$\tan\alpha + \tan 2\beta = \dfrac{1}{7} + \dfrac{3}{4} = \dfrac{4}{28} + \dfrac{21}{28} = \dfrac{25}{28}.$$

Then the denominator:

$$1 - \tan\alpha\,\tan 2\beta = 1 - \left(\dfrac{1}{7}\right)\left(\dfrac{3}{4}\right) = 1 - \dfrac{3}{28} = \dfrac{25}{28}.$$

Hence

$$\tan(\alpha + 2\beta) = \dfrac{\dfrac{25}{28}}{\dfrac{25}{28}} = 1.$$

Hence, the correct answer is Option 1.