Exams › JEE Advanced › Maths

Find the value of the product tan(60 deg)*tan(42 deg)*tan(66 deg)*tan(78 deg).

1. tan 6 deg
2. 1
3. sqrt(3)
4. tan 60 deg

Correct answer: tan 6 deg

Solution

Applying tan theta*tan(60-theta)*tan(60+theta) = tan 3 theta repeatedly, the product tan60*tan42*tan66*tan78 collapses to tan6 deg.

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 →