Animated Solution for Mathematics - Differentiation: If $2y={\left({\cot }^{-1}\left(\frac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x}\right)\right)}^2,x\in (0,\frac{\pi }{2})$, then $\frac{dy}{dx}$ is equal to :

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of inverse trigonometry.

You are presented with the equation $2y={\left({\cot }^{-1}\left(\frac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x}\right)\right)}^2$.

Your instinct might be to reach for the chain rule and start differentiating immediately. But pause. In the world of competitive mathematics, brute force is rarely the intended path. Let us look for the elegance hidden beneath the surface.

Trigonometric Surgery

Look closely at the fraction inside the ${\cot }^{-1}$ function: $\frac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x}$. This is a classic structure.

Whenever you see a linear combination of $\sin x$ and $\cos x$ in a fraction, your first instinct should be to divide the numerator and the denominator by $\cos x$. Watch what happens:

Suddenly, the complexity evaporates. We have transformed a bulky fraction into a form that screams for a trigonometric identity.

Recall that $\sqrt{3}$ is simply $\tan (\frac{\pi }{3})$. Substituting this in, we get:

The Identity Bridge

Does this look familiar? It is the exact expansion of the compound angle formula: $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$.

By setting $A=\frac{\pi }{3}$ and $B=x$, our entire fraction collapses into a single, beautiful term: $\tan (\frac{\pi }{3}+x)$.

Now, our original equation looks much friendlier: $2y={\left({\cot }^{-1}\left(\tan \left(\frac{\pi }{3}+x\right)\right)\right)}^2$. We are making progress!

The Inverse Dance

We have a ${\cot }^{-1}$ on the outside and a $\tan$ on the inside. They do not cancel directly.

To bridge this gap, we use the complementary angle identity: $\tan \theta =\cot (\frac{\pi }{2}-\theta )$. Applying this to our angle $\theta =\frac{\pi }{3}+x$, we get:

Since the angle is within the principal domain, the ${\cot }^{-1}$ and $\cot$ neutralize each other. We are left with the angle: $\frac{\pi }{2}-\frac{\pi }{3}-x$, which simplifies to $\frac{\pi }{6}-x$.

Our intimidating equation has now become the simple algebraic expression: $2y={\left(\frac{\pi }{6}-x\right)}^2$.

The Final Calculus

Now, and only now, do we perform the differentiation. Differentiating both sides with respect to $x$ using the chain rule:

The derivative of $(\frac{\pi }{6}-x)$ is simply $-1$. Thus, we have:

Dividing by $2$, we arrive at our final result: $\frac{dy}{dx}=x-\frac{\pi }{6}$.

This problem is a masterclass in why we simplify before we differentiate. If you had jumped straight into the derivative, you would have been lost in a forest of quotient rules and chain rules.

By taking a moment to perform trigonometric surgery, we turned a mountain into a molehill. Keep this in your toolkit: whenever you see inverse trig, look for the identity that simplifies the interior first. You have got this!