JEE Main 2025 · 4 April · morning · Single correct · +4 / −1
Considering the principal values of the inverse trigonometric functions, sin ⁻¹(√3/2 x+1/2 √(1-x²)),-1/2< x<1/√2, is equal to
- (a) (-5 π)/6-sin ⁻¹ x
- (b) (5 π)/6-sin ⁻¹ x
- (c) π/6+sin ⁻¹ x
- (d) π/4+sin ⁻¹ x
Answer and solution
Answer (c) π/6+sin ⁻¹ x
Solution
Maths sir · 4 steps
Put x = sin θ, spot sin(θ + π/6), then check the range.
- Substitute. Let θ = sin⁻¹ x, so x = sin θ. Since θ is a principal value, cos θ ≥ 0, so √(1-x²) = cos θ.
- Spot the formula. Read √3/2 as cos π/6 and 1/2 as sin π/6. The bracket becomes the expansion of sin(A+B).
√3/2 sin θ + 1/2 cos θ = sin θ cos π/6 + cos θ sin π/6 = sin(θ + π/6)
- Check the range. -1/2 < x < 1/√2 gives -π/6 < θ < π/4, so 0 < θ + π/6 < 5π/12. This lies inside [-π/2, π/2], so sin⁻¹ undoes sin exactly.
- So. The expression equals θ + π/6 = π/6 + sin⁻¹ x, which is option (c).
The idea to keep. sin⁻¹(sin α) = α only when α lies in [-π/2, π/2]. After you combine the angles, always check the range before you cancel.
Answer from the NTA final answer key. Solution by PrepWiser.