Exams › JEE Advanced › Maths

Find the value of x satisfying sin⁻¹(tan(pi/4)) - sin⁻¹(sqrt(3/x)) - pi/6 = 0.

1. x = 2
2. x = -4
3. x = 4
4. x = 3

Correct answer: x = 4

Solution

Since tan(pi/4) = 1, sin⁻¹(1) = pi/2. The equation becomes pi/2 - sin⁻¹(sqrt(3/x)) = pi/6, so sin⁻¹(sqrt(3/x)) = pi/2 - pi/6 = pi/3. Taking sine: sqrt(3/x) = sin(pi/3) = sqrt(3)/2. Squaring: 3/x = 3/4, hence x = 4.

Related JEE Advanced Maths questions

• Identify the periodic function among the following:
• Given that α, β, γ, and θ are the smallest positive angles in increasing order for which their sine values equal a positive constant k, what is the result of 4 sin(α/2) + 3 sin(β/2) + 2 sin(γ/2) + sin(θ/2)?
• Consider the function fn(θ) = tan(θ/2)(1 + sec²θ)(1 + sec⁴θ)... (1 + sec²ⁿθ). Which of the following is true?
• Given that (a−b)sin(θ+ϕ) = (a+b)sin(θ−ϕ) and a tan(θ/2) − b tan(ϕ/2) = c, which of the following holds true?
• In a right-angled triangle ABC where ∠B = 90° and the sum of the two legs a and b equals 4, the triangle's area is maximized when ∠C is:
• If the altitudes p₁, p₂, and p₃ of a triangle enclose a circle with a diameter of 4/3 units, what is the minimum possible value of p₁ + p₂ + p₃?

⚔️ Practice JEE Advanced Maths free + battle 1v1 →