Animated Solution for Mathematics - Trigonometry: If $\frac{\sqrt{2}\sin \alpha }{\sqrt{1+\cos 2\alpha }}=\frac{1}{7}$ and $\sqrt{\frac{1-\cos 2\beta }{2}}=\frac{1}{\sqrt{10}}$, $\alpha ,\beta \in (0,\frac{\pi }{2})$, then $\tan (\alpha +2\beta )$ is equal to _________ .

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a problem that, at first glance, looks like a tangled mess of square roots and double angles.

But as we peel back the layers, you will see the elegance hidden within. This is not just about solving for a value; it is about recognizing the patterns that govern trigonometry.

Decoding the Identities

Let us look at our starting equations. We are given:

and

The first thing that should jump out at you is the presence of $1+\cos 2\alpha$ and $1-\cos 2\beta$. These are classic triggers for the double angle identities.

Recall that $1+\cos 2\theta =2{\cos }^2\theta$ and $1-\cos 2\theta =2{\sin }^2\theta$. By substituting these, we transform the intimidating square roots into something much more manageable.

The denominator of our first equation becomes $\sqrt{2{\cos }^2\alpha }$, and the numerator of our second becomes $\sqrt{2{\sin }^2\beta }$.

The Geometry of $\alpha$ and $\beta$

With the identities applied, the equations simplify beautifully. For $\alpha$, we get:

The $\sqrt{2}$ terms cancel out, leaving us with $\tan \alpha =\frac{1}{7}$.

For $\beta$, we have $\sqrt{\frac{2{\sin }^2\beta }{2}}=\frac{1}{\sqrt{10}}$, which simplifies to $\sin \beta =\frac{1}{\sqrt{10}}$. Since $\beta$ is in the first quadrant, we know $\sin \beta$ is positive.

Now, imagine a right-angled triangle for $\beta$. If the opposite side is $1$ and the hypotenuse is $\sqrt{10}$, then by the Pythagorean theorem, the adjacent side is $\sqrt{(\sqrt{10}{)}^2-1^2}=\sqrt{9}=3$. Thus, $\tan \beta =\frac{1}{3}$.

The Double Angle Bridge

We are now armed with $\tan \alpha =\frac{1}{7}$ and $\tan \beta =\frac{1}{3}$. Our goal is to find $\tan (\alpha +2\beta )$.

We have $\alpha$, but we need $2\beta$. This is where the double angle formula for tangent comes into play: $\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$.

Substituting $\tan \beta =\frac{1}{3}$, we get:

The Grand Finale

Finally, we use the compound angle formula: $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$. Setting $A=\alpha$ and $B=2\beta$, we have:

Plugging in our values, $\tan \alpha =\frac{1}{7}$ and $\tan 2\beta =\frac{3}{4}$, we get:

Look at that! The numerator and denominator are identical. The result is $1$.

This is the beauty of mathematics—when you trust the process and the identities, even the most complex-looking problems collapse into simple, elegant truths. Keep practicing, and you will start to see these patterns everywhere!