Animated Solution for Mathematics - Trigonometry: Let $\cos (\alpha +\beta )=4/5$ and $\sin (\alpha -\beta )=5/13$, where $0\le \alpha ,\beta \le \pi /4$. Then $\tan 2\alpha =$

A JEE Masterclass

Welcome, future engineer! Today, we are going to dismantle a classic trigonometry problem that tests not just your memory of formulas, but your ability to see the hidden structure within an equation.

When you look at $\cos (\alpha +\beta )=4/5$ and $\sin (\alpha -\beta )=5/13$, your first instinct might be to panic about finding $\alpha$ and $\beta$. But stop. Take a breath.

In the world of JEE Advanced, we don't solve for variables when we can solve for the structure.

The Substitution Strategy

The secret to this problem lies in a simple, yet profound, substitution. Let us define two new angles: $A=\alpha +\beta$ and $B=\alpha -\beta$.

Suddenly, the problem transforms. We are no longer dealing with complex sums and differences; we are dealing with simple, clean ratios: $\cos A=4/5$ and $\sin B=5/13$.

This is the first step in the 'JEE Mindset'—simplifying the landscape before you start the journey.

Geometric Intuition

Now, let's visualize these angles. We know $\cos A=4/5$.

Imagine a right-angled triangle where the base is $4$ and the hypotenuse is $5$. By the Pythagorean theorem, the perpendicular must be $\sqrt{5^2-4^2}=3$. Thus, $\tan A=3/4$.

Similarly, for angle $B$, we have $\sin B=5/13$. In this triangle, the perpendicular is $5$ and the hypotenuse is $13$. The base is $\sqrt{13^2-5^2}=12$. Therefore, $\tan B=5/12$.

We have successfully converted our given information into the language of tangents.

The Elegant Bridge

Now, look at the target: $\tan 2\alpha$. How does this relate to our $A$ and $B$?

If you add them, $A+B=(\alpha +\beta )+(\alpha -\beta )=2\alpha$. This is the 'Aha!' moment.

Finding $\tan 2\alpha$ is exactly the same as finding $\tan (A+B)$. We have built a bridge between the given data and the final answer.

The Final Calculation

We use the compound angle identity:

Substituting our values, we get:

Let's handle the numerator: $3/4+5/12=9/12+5/12=14/12=7/6$.

Now the denominator: $1-15/48=(48-15)/48=33/48$.

Finally, we divide:

And there it is! The elegance of the final result, $56/33$, is the reward for your disciplined approach. Remember, in trigonometry, always look for the hidden sum or difference. It is the key that unlocks the most difficult problems.