Animated Solution for Mathematics - Inverse Trigonometric Functions: Match List I with List II:

Welcome, aspiring engineers. Today, we are not just solving a math problem; we are engaging in a high-stakes dance with trigonometry. The 'Match the Following' format is a staple of the JEE Advanced examination.

It tests not just your ability to calculate, but your ability to recognize patterns, maintain composure under pressure, and execute algebraic maneuvers with surgical precision. Let us dissect this problem, piece by piece, and uncover the elegance hidden within these expressions.

The Triangle Visualization

When you look at the expression in part (P), it is natural to feel a sense of dread. It looks like a chaotic mess of inverse functions and squares. But remember the golden rule of JEE: Substitution is your best friend.

We start by letting ${\tan }^{-1}y=\theta$. This implies $\tan \theta =y$. Now, visualize a right-angled triangle. If the opposite side is $y$ and the adjacent side is $1$, the hypotenuse must be $\sqrt{1+y^2}$.

From this simple geometric construct, we can immediately write:

Substituting these into the numerator, we get:

Now, we do the same for the denominator using ${\sin }^{-1}y=\varphi$. The denominator simplifies to $\frac{1}{y\sqrt{1-y^2}}$. When we put it all together inside the square root, the terms cancel with such satisfying precision that you realize the entire expression collapses to $1$.

The Power of Squaring

Next, we face a system of equations: $\cos x+\cos y=-\cos z$ and $\sin x+\sin y=-\sin z$. Many students try to solve for $x$ and $y$ individually, but that is a path to frustration. Instead, we use the 'Squaring and Adding' technique.

By squaring both equations, we invoke the most fundamental identity in trigonometry: ${\sin }^2\theta +{\cos }^2\theta =1$. When we add the squared equations, the cross-terms $2\cos x\cos y$ and $2\sin x\sin y$ combine to form $2\cos (x-y)$.

The right side becomes ${\cos }^{2}z+{\sin }^{2}z$, which is simply $1$. We are left with:

Using the double-angle formula $\cos 2\theta =2{\cos }^2\theta -1$, we find the half-angle value. It is a beautiful example of how symmetry in equations allows us to bypass the variables entirely.

Pattern Recognition

Part (R) is where your ability to spot identities is tested. We are given an equation involving $\cos (\frac{\pi }{4}-x)$ and $\cos (\frac{\pi }{4}+x)$. The moment you see a difference of cosines, you should think of the identity:

By grouping the terms and applying this identity, the left side of the equation transforms into $\sqrt{2}\sin x\cos 2x$. On the right side, we expand $\sin 2x$ as $2\sin x\cos x$ and $\sec x$ as $\frac{1}{\cos x}$.

The $\cos x$ terms cancel out, leaving us with a much simpler equation. Assuming $\sin x eq 0$ and $\cos x - \sin x eq 0$, we eventually arrive at $\cos x+\sin x=\sqrt{2}$. This is a classic condition that holds when $x=\frac{\pi }{4}$, leading us directly to $\sec x=\sqrt{2}$.

The Final Boss

Finally, we arrive at part (S). We have $\cot ({\sin }^{-1}\sqrt{1-x^2})=\sin ({\tan }^{-1}(x\sqrt{6}))$. The left side is a classic identity: ${\sin }^{-1}\sqrt{1-x^2}$ is equivalent to ${\cos }^{-1}x$.

Thus, $\cot ({\cos }^{-1}x)$ becomes:

On the right side, we use the same triangle method we used in part (P). Letting ${\tan }^{-1}(x\sqrt{6})=\beta$, we find:

Equating the two sides and squaring both sides to eliminate the radicals, we get:

A quick cross-multiplication and rearrangement gives $12x^2=5$, or $x=\frac{1}{2}\sqrt{\frac{5}{3}}$.

We have navigated through four distinct mathematical landscapes. We used triangle visualization, symmetry, identity recognition, and algebraic manipulation.

The key takeaway for your JEE journey is this: Do not fear the complexity. Every complex expression is built from simple, fundamental truths. Your job is to peel back the layers, one identity at a time, until the truth reveals itself.