If (3 + i sin θ)/(4 - i cos θ), θ ∈ [0, 2θ], is a real number, then an argument of
sinθ + icosθ is :

Solution

Maths sir · 3 steps

Make the imaginary part zero to get tan θ, then find the quadrant of sin θ + i cos θ.

1. Rationalise. Multiply the top and bottom by 4 + i cos θ. The denominator becomes 16 + cos² θ, which is real and positive. So the fraction is real exactly when the imaginary part of the new numerator is zero.

   (3 + i sin θ)(4 + i cos θ) = (12 - sin θ cos θ) + i (3 cos θ + 4 sin θ)

2. Find the angle. Set 3 cos θ + 4 sin θ = 0, which gives tan θ = -3/4. So sin θ and cos θ have opposite signs, with |sin θ| = 3/5 and |cos θ| = 4/5.

   Case for θ	sin θ + i cos θ	Quadrant of the point	Argument
   second quadrant	3/5 - 4/5i	fourth	-tan⁻¹(4/3), not an option
   fourth quadrant	-3/5 + 4/5i	second	π - tan⁻¹(4/3)

3. So. In the second case the point has a negative real part and a positive imaginary part, with imaginary/|real| = 4/3. That puts its argument at π - tan⁻¹(4/3), which is option (d).

The idea to keep. For the argument of x + iy, first fix the quadrant from the signs of x and y, then use tan⁻¹|y/x|. Here the real part is sin θ, not cos θ.

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