Exams › JEE Main › Maths

Prove that (sin²(A) - sin²(B)) / (sin(A)cos(A) - sin(B)cos(B)) = tan(A + B). What is the value of the left-hand expression?

1. tan(A + B)
2. tan(A - B)
3. cot(A + B)
4. tan(A)tan(B)

Correct answer: tan(A + B)

Solution

Numerator = sin(A+B)sin(A-B); denominator (after doubling) sin2A - sin2B = 2cos(A+B)sin(A-B). The ratio reduces to sin(A+B)/cos(A+B) = tan(A+B).

Related JEE Main Maths questions

• Find the period of the function |sin³(x/2)| + |cos⁵(x/5)|.
• For which values of a does the equation sin⁴ x + cos⁴ x = a admit at least one solution?
• For n ∈ N, let fₙ(θ)=tan ((θ)/(2)) (1+secθ)(1+sec 2θ)(1+sec 4θ)⋯(1+sec 2ⁿθ). Which of the following is true?
• Simplify the following expression: (cos 6x + 6cos 4x + 15cos 2x + 10)/(cos 5x + 5cos 3x + 10cos x). What does it equal?
• For angles α, β, γ all lying in the interval (0, π/2), the value of (sin(α+β+γ))/(sinα+sinβ+sinγ) is
• Evaluate the product (1 + cos(π/10))(1 + cos(3π/10))(1 + cos(7π/10))(1 + cos(9π/10)).

⚔️ Practice JEE Main Maths free + battle 1v1 →