Q48 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics

If $\sqrt{5}-i \sqrt{15}=r(\cos \theta+i \sin \theta),-\pi \lt \theta \lt \pi$, then $r^2\left(\sec \theta+3 \operatorname{cosec}^2 \theta\right)=$

  • A
    40
  • B
    60
  • Correct answer
    C
    120
  • D
    180

Solution

$\sqrt{5}-i \sqrt{15}=2 \sqrt{5}\left(\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)$ $\begin{aligned} & \cos \theta=\frac{1}{2} \Rightarrow \sec \theta=2 \\ & \sin \theta=\frac{-\sqrt{3}}{2} \Rightarrow \operatorname{cosec} \theta=\frac{-2}{\sqrt{3}} \\ & r=2 \sqrt{5} \\ & r^2\left(\sec \theta+3 \operatorname{cosec}^2 \theta\right)=20 \times\left(2+3 \times \frac{4}{3}\right)=120 .\end{aligned}$

Hence, the correct answer is option: 'C'.

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