If (sin⁻¹ x)² - (cos⁻¹ x)² = a; 0

Q: Why can't we just solve for $x$ directly by isolating the inverse functions?

Solving for $x$ directly would lead to a messy transcendental equation involving both $\sin^{-1} x$ and $\cos^{-1} x$. In JEE Advanced, we prioritize structural symmetry over brute force. By using identities, we transform the problem into a linear relationship, which is much more elegant and less prone to calculation errors.

Q: Is the identity $2\cos^{-1} x = \cos^{-1}(2x^2 - 1)$ always true?

Great catch! This identity is valid specifically for $x \in [0, 1]$. Since the problem explicitly states $0 < x < 1$, we are safely within the domain. Always remember to check the domain constraints in inverse trigonometry problems—it is a classic trap!

Q: Why did we convert $\sin^{-1} x$ to $\cos^{-1} x$ instead of the other way around?

It is a matter of convenience. We needed to reach an expression involving $2x^2 - 1$, which is the argument of the double angle formula for $\cos^{-1}$. Converting everything to $\cos^{-1} x$ aligns our equation perfectly with the target expression, making the final step much smoother.

Question

If $(\sin^{-1} x)^2 - (\cos^{-1} x)^2 = a; 0 < x < 1, a 
eq 0$, then the value of $2x^2 - 1$ is :

Options:

(A) $\cos\left(\frac{4a}{\pi}ight)$

(B) $\sin\left(\frac{2a}{\pi}ight)$

(C) $\cos\left(\frac{2a}{\pi}ight)$

(D) $\sin\left(\frac{4a}{\pi}ight)$

Sigma Percentile · Accepted Answer

### The Beauty of Algebraic Symmetry

Welcome, future engineer! Today, we are going to peel back the layers of an inverse trigonometry problem that, at first glance, might seem like a tangled mess of functions.

But I want you to take a deep breath. In JEE Advanced, the most intimidating problems are often just simple identities wearing a disguise. Let's embark on this journey together.

### Phase 1: The Algebraic Dance

We start with the equation $(\sin^{-1} x)^2 - (\cos^{-1} x)^2 = a$. Your intuition might scream, "Should I find $x$?" No! That is the trap.

Instead, look at the structure. It is a perfect difference of squares, $A^2 - B^2$, which we know expands to $(A + B)(A - B)$.

By treating $A = \sin^{-1} x$ and $B = \cos^{-1} x$, we factorize our equation into:

$$(\sin^{-1} x + \cos^{-1} x)(\sin^{-1} x - \cos^{-1} x) = a$$

Suddenly, the complexity begins to dissolve.

### Phase 2: The Bridge of Identities

Now, look at that first bracket: $(\sin^{-1} x + \cos^{-1} x)$. Does it ring a bell? It is the bedrock of inverse trigonometry!

We know that for any $x \in [-1, 1]$, the following identity holds:

$$\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$$

Since our constraint is $0 < x < 1$, we are perfectly safe to use this. Substituting this into our equation, we get:

$$\frac{\pi}{2}(\sin^{-1} x - \cos^{-1} x) = a$$

By multiplying both sides by $\frac{2}{\pi}$, we isolate the difference:

$$\sin^{-1} x - \cos^{-1} x = \frac{2a}{\pi}$$

We have successfully reduced a quadratic-looking equation into a linear one.

### Phase 3: The Final Transformation

We are almost there, but we have two different functions. Let's unify them.

Using the identity $\sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x$, our equation becomes:

$$\left( \frac{\pi}{2} - \cos^{-1} x ight) - \cos^{-1} x = \frac{2a}{\pi}$$

Combining the terms, we arrive at:

$$\frac{\pi}{2} - 2\cos^{-1} x = \frac{2a}{\pi}$$

Rearranging this gives us:

$$2\cos^{-1} x = \frac{\pi}{2} - \frac{2a}{\pi}$$

### The Grand Finale

Here is the masterstroke. We need to find $2x^2 - 1$. Recall the double angle property:

$$2\cos^{-1} x = \cos^{-1}(2x^2 - 1)$$

Substituting this in, we get:

$$\cos^{-1}(2x^2 - 1) = \frac{\pi}{2} - \frac{2a}{\pi}$$

To free our target expression, we take the cosine of both sides:

$$2x^2 - 1 = \cos\left( \frac{\pi}{2} - \frac{2a}{\pi} ight)$$

Using the allied angle identity $\cos(\frac{\pi}{2} - 	heta) = \sin 	heta$, we finally arrive at the result:

**$$2x^2 - 1 = \sin\left( \frac{2a}{\pi} ight)$$**

See how the pieces fit? We didn't fight the math; we guided it. Keep this mindset, and you will conquer any problem the exam throws at you!

Animated Solution for Mathematics - Inverse Trigonometric Functions: If $(sin^{- 1} x)^{2} - (cos^{- 1} x)^{2} = a; 0 < x < 1, a \neq = 0$ , then the value of $2 x^{2} - 1$ is :

Select Answer:

Visualized Solution

The Given Equation

Factorizing $A^{2} - B^{2}$

The $\frac{π}{2}$ Identity

Substituting the Identity

Simplifying the Equation

Expressing in terms of $cos^{- 1} x$

Combining Like Terms

Isolating $2 cos^{- 1} x$

The Double Angle Formula

Applying the Formula

Taking Cosine on Both Sides

Using Allied Angles

The Sigma Insight: Properties of Inverse Trigonometric Functions

The Beauty of Algebraic Symmetry

Phase 1

Phase 2

Phase 3

The Grand Finale

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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 (sin−1x)2−(cos−1x)2=a;0<x<1,a=0, then the value of 2x2−1 is :

Select Answer:

Visualized Solution

The Given Equation

Factorizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2−B2

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2π​ Identity