Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy
sin ⁻¹( 3x/5 ) + sin ⁻¹( 4x/5 ) = sin ⁻¹x is equal to :

Solution

Maths sir · 5 steps

Take sine of both sides and factor out the common x. Do not divide by it, because x = 0 is a root too.

1. Take sine. Let A = sin⁻¹ 3x/5 and B = sin⁻¹ 4x/5. Principal values lie in [-π/2, π/2], so cos A = √(25-9x²)/5 and cos B = √(25-16x²)/5, both non-negative. Also |x| ≤ 1, because sin⁻¹ x must exist.
2. Expand sin(A + B). From sin A cos B + cos A sin B = x you get (3x√(25-16x²) + 4x√(25-9x²))/25 = x. Every term has a factor x, so either x = 0 or the bracket below holds.

   3√(25-16x²) + 4√(25-9x²) = 25

3. Solve the bracket. The left side only gets smaller as x² grows. It equals 35 at x² = 0 and 3 × 3 + 4 × 4 = 25 at x² = 1. So it reaches 25 only at x² = 1, which gives x = ± 1.
4. Check directly. For x = 1, the angles sin⁻¹ 3/5 and sin⁻¹ 4/5 are complementary, because cos(sin⁻¹ 3/5) = 4/5. Their sum is π/2 = sin⁻¹ 1. Both sides are odd functions of x, so x = -1 works too, and x = 0 gives 0 = 0.
5. So. The solutions are x = 0, 1, -1. That is 3 real values, option (c).

The idea to keep. When x appears as a factor on both sides, take it out as a common factor and never cancel it. Then check every root in the original equation, since principal values limit the sum of the angles.

Answer from the NTA final answer key. Solution by PrepWiser.