If (sin⁻¹x)/a = (cos⁻¹x)/b = (tan⁻¹y)/c; 0

Q: Wait, why can we just add the numerators and denominators? Is that always allowed?

That is a fantastic question! You are referring to the property of equal ratios (also known as the componendo rule). If $\frac{A}{a} = \frac{B}{b} = \frac{C}{c} = k$, then it is mathematically true that $\frac{A+B}{a+b} = k$. We are essentially scaling the ratios to find a common ground that matches our target expression. It is a powerful tool in JEE problems to bridge the gap between given variables and the final answer.

Q: I got $\cos(2\theta)$, but how do I convert that to $y$ without finding $\theta$?

You are on the right track! You don't need to find $\theta$ explicitly. Since we defined $\theta = \tan^{-1}y$, we know $\tan\theta = y$. We can use the double-angle formula for cosine in terms of tangent: $\cos(2\theta) = \frac{1-\tan^2\theta}{1+\tan^2\theta}$. By substituting $y$ for $\tan\theta$, you get the answer directly in terms of $y$ without ever needing to calculate the inverse tangent.

Q: What if $x$ was outside the range $(0, 1)$? Would the identity still hold?

The identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ is valid for all $x$ in the domain $[-1, 1]$. The problem restricts $x$ to $(0, 1)$ to ensure we are working within the principal value branches where these functions are well-defined and continuous. If $x$ were outside this range, the inverse functions would not be defined in the standard real-valued sense, so the identity would not apply. Always check your domain constraints!

Question

If $\frac{\sin^{-1}x}{a} = \frac{\cos^{-1}x}{b} = \frac{	an^{-1}y}{c}$; $0 < x < 1$, then the value of $\cos\left(\frac{\pi c}{a+b}ight)$ is:

Options:

(A) $\frac{1-y^2}{2y}$

(B) $\frac{1-y^2}{1+y^2}$

(C) $1$

(D) $\frac{y}{\sqrt{1-y^2}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are dissecting a problem that, at first glance, looks like a tangled mess of inverse trigonometric functions.

We are given the relation:
$$\frac{\sin^{-1}x}{a} = \frac{\cos^{-1}x}{b} = \frac{	an^{-1}y}{c}$$

Our mission is to find the value of $\cos\left(\frac{\pi c}{a+b}ight)$. When you see a problem like this, do not panic; instead, look for the structural hints. The target expression contains $\frac{\pi c}{a+b}$, and the presence of $a+b$ in the denominator is a massive clue to utilize the property of equal ratios.

### The Master Key

Before we dive into the algebra, let us recall the fundamental identity that acts as our master key:
$$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$$

This identity is valid for $x \in [0, 1]$, which is the domain provided in our problem. This constant $\frac{\pi}{2}$ is the bridge we need to unlock the entire expression.

### The Art of Combination

Let us apply the property of equal ratios. If $\frac{A}{a} = \frac{B}{b}$, then both are equal to $\frac{A+B}{a+b}$.

Applying this to our first two terms, we get:
$$\frac{\sin^{-1}x}{a} = \frac{\cos^{-1}x}{b} = \frac{\sin^{-1}x + \cos^{-1}x}{a+b}$$

Now, substitute our master key: the numerator becomes $\frac{\pi}{2}$. Thus, our combined ratio is:
$$\frac{\pi}{2(a+b)}$$

We have successfully created the structural term $\frac{\pi}{a+b}$ required for our target.

### The Geometric Bridge

Now, we equate this to the third term:
$$\frac{\pi}{2(a+b)} = \frac{	an^{-1}y}{c}$$

Rearranging this, we find:
$$\frac{\pi c}{a+b} = 2	an^{-1}y$$

Let us pause and appreciate this. We have reduced an intimidating expression into the cosine of $2	an^{-1}y$. Let $	heta = 	an^{-1}y$, which implies $	an	heta = y$.

### The Elegant Conclusion

We recall the double-angle formula for cosine in terms of tangent:
$$\cos(2	heta) = \frac{1-	an^2	heta}{1+	an^2	heta}$$

Since $	an	heta = y$, we substitute $y$ into this formula to obtain the final result:
$$\cos\left(\frac{\pi c}{a+b}ight) = \mathbf{\frac{1-y^2}{1+y^2}}$$

This is the essence of JEE Advanced mathematics: identifying the structure, applying the right identity, and watching the complexity vanish. Keep practicing this structural thinking, and you will find that even the most intimidating problems have a simple, elegant soul waiting to be revealed.

Animated Solution for Mathematics - Inverse Trigonometric Functions: If $\frac{s i n ^{- 1} x}{a} = \frac{c o s ^{- 1} x}{b} = \frac{t a n ^{- 1} y}{c}$ ; $0 < x < 1$ , then the value of $cos (\frac{π c}{a + b})$ is:

Select Answer:

Visualized Solution

Identify the Given Ratios

Recall the Fundamental Identity

Apply the Property of Ratios

Substitute the Identity Value

Equate to the Third Ratio

Visualize the Angle $θ$

Assign Triangle Sides

Apply the Cosine Function

Recall the Double Angle Formula

Final Substitution and Conclusion

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

The Master Key

The Art of Combination

The Geometric Bridge

The Elegant Conclusion

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Animated Solution for Mathematics - Inverse Trigonometric Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If asin−1x​=bcos−1x​=ctan−1y​; 0<x<1, then the value of cos(a+bπc​) is: