cosec18° is a root of the equation :

Solution

Maths sir · 4 steps

Get the quadratic for sin 18° first, then put s = 1/x to flip it into one for cosec.

1. Key trick for 18°. Let θ = 18°. Then 5θ = 90°, so 2θ = 90° - 3θ, which gives sin 2θ = cos 3θ.
2. Quadratic in sine. Expand: 2 sin θ cos θ = 4 cos³ θ - 3 cos θ. Divide by cos θ ≠ 0 to get 2 sin θ = 4 cos² θ - 3 = 1 - 4 sin² θ.

   4 sin² 18° + 2 sin 18° - 1 = 0

3. Switch to cosec. Put sin 18° = 1/x, where x = cosec 18°. This gives 4/x² + 2/x - 1 = 0. Multiply by x² to get 4 + 2x - x² = 0, that is x² - 2x - 4 = 0.
4. So. As a check, sin 18° = (√5-1)/4, so cosec 18° = √5+1. Then (√5+1)² - 2(√5+1) - 4 = 6 + 2√5 - 2√5 - 2 - 4 = 0. The answer is (d).

The idea to keep. If you know the equation for sin θ, the equation for cosecθ is the same equation with its coefficients written in reverse order. And remember 5 × 18° = 90°.

Answer from the NTA final answer key. Solution by PrepWiser.