Animated Solution for Mathematics - Trigonometry: If $\tan \alpha =\frac{m}{m+1}$ and $\tan \beta =\frac{1}{2m+1}$, find the possible values of $(\alpha +\beta )$.

Imagine you are standing at the edge of a complex algebraic landscape. You are given two angles, $\alpha$ and $\beta$, defined by the parameter $m$.

At first glance, the expressions $\tan \alpha =\frac{m}{m+1}$ and $\tan \beta =\frac{1}{2m+1}$ seem to suggest that the sum of these angles will change wildly as $m$ changes.

But in the world of JEE Advanced, appearances can be deceiving. Today, we are going to uncover a hidden truth: that the sum $(\alpha +\beta )$ is actually a constant, independent of $m$.

The Compound Angle Formula

To find the sum of two angles when we only know their tangents, we need a bridge. That bridge is the compound angle identity:

This formula is our most powerful tool. It transforms the sum of two angles into a ratio of their tangents.

It is the key that will allow us to combine these two seemingly disparate expressions into one unified result.

Substitution and Simplification

Now, let us dive into the algebra. We substitute our given values into the formula:

I know this looks intimidating. It is a complex fraction, a fraction within a fraction. But take a deep breath.

Let us tackle the numerator first. To add $\frac{m}{m+1}$ and $\frac{1}{2m+1}$, we find a common denominator, which is $(m+1)(2m+1)$.

The numerator becomes $m(2m+1)+1(m+1)$. Expanding this, we get $2m^2+m+m+1$, which simplifies beautifully to $2m^2+2m+1$.

Now, let us look at the denominator: $1-\frac{m}{(m+1)(2m+1)}$. Again, we use the common denominator $(m+1)(2m+1)$.

The expression becomes:

Expanding the product $(m+1)(2m+1)$, we get $2m^2+3m+1$. Subtracting $m$ from this, we are left with $2m^2+2m+1$.

The Moment of Cancellation

Look closely at what we have achieved. The numerator is $\frac{2m^2+2m+1}{(m+1)(2m+1)}$ and the denominator is also $\frac{2m^2+2m+1}{(m+1)(2m+1)}$.

When we divide these two identical expressions, they cancel out completely! We are left with:

This is the 'Aha!' moment. All the complexity of $m$ has vanished, leaving us with a clean, elegant constant.

The General Solution

We have arrived at $\tan (\alpha +\beta )=1$. We know that $\tan (\frac{\pi }{4})=1$.

However, we must remember that the tangent function is periodic. It repeats its values every $\pi$ radians.

Therefore, the general solution for $(\alpha +\beta )$ is:

where $n$ is any integer. This result is profound.

It tells us that no matter what value $m$ takes, the sum of these two angles will always be an odd multiple of $\frac{\pi }{4}$ (specifically, $\frac{\pi }{4},\frac{5\pi }{4},-\frac{3\pi }{4}$, etc.).

You have just navigated a complex problem and found the underlying simplicity. Keep this spirit of curiosity and persistence, and you will conquer any problem the JEE throws at you.