If the real part of the complex number (1 - cos θ + 2i sin θ )⁻¹ is 1/5 for θ ∈ (0,π ), then the value of the integral ∫₀^θ sin x dx is equal to:

Solution

Maths sir · 4 steps

Rationalise, then write sin² θ as (1-cos θ)(1+cos θ) so a factor cancels.

1. Real part. For z = a + ib, the real part of 1/z is a/(a²+b²). Here a = 1-cos θ and b = 2 sin θ, so b² = 4 sin² θ = 4(1-cos θ)(1+cos θ).

   (1-cos θ)/((1-cos θ)² + 4(1-cos θ)(1+cos θ)) = 1/((1-cos θ)+4(1+cos θ)) = 1/(5+3 cos θ)

2. Cancel safely. For θ ∈ (0,π), cos θ ≠ 1, so 1-cos θ ≠ 0 and the cancellation is allowed. Setting 1/(5+3 cos θ) = 1/5 gives cos θ = 0, so θ = π/2.
3. Evaluate integral. ∫₀^θ sin x dx = [-cos x]₀^θ = 1-cos θ. With cos θ = 0, this is 1.
4. So. The integral equals 1, which is option (a).

The idea to keep. When a complex number has 1-cos θ and sin θ in it, write sin² θ = (1-cos θ)(1+cos θ). A common factor usually cancels and the algebra becomes linear.

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