Exams › JEE Main › Maths

Show that cosec⁶(alpha) - cot⁶(alpha) = 3*cosec²(alpha)*cot²(alpha) + 1. As an identity, which equivalent simplified form does the left side reduce to?

1. 3*cosec²(alpha)*cot²(alpha) + 1
2. 3*cosec²(alpha)*cot²(alpha) - 1
3. cosec²(alpha)*cot²(alpha) + 1
4. 3*cosec²(alpha)*cot²(alpha) + 3

Correct answer: 3*cosec²(alpha)*cot²(alpha) + 1

Solution

Treating cosec² and cot² as a and b, the difference of cubes factors using a - b = 1, leaving 3ab + 1.

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 →