Q49 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics
If $\left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^{2020}+\left(\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta+i \sin \theta}\right)^{2021}$ $=x+i y$, then the value of $x+y$ at $\theta=\frac{\pi}{2}$ is
- Correct answerA2
- B1
- C-1
- D2020
Solution
At $\theta=\frac{\pi}{2}$
$\left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^{2020}+\left(\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta+i \sin \theta}\right)^{2021}$
$=\left(\frac{0+i \cdot 1}{1+i \cdot 0}\right)^{2020}+\left(\frac{1+0+i}{1-0+i}\right)^{2021}$
$=(i)^{2020}+\left(\frac{1+i}{1+i}\right)^{2021}=1+(1)^{2021}=1+1$
$=2=x+i y \Rightarrow x=2, y=0$
$\therefore x+y=2+0=2$
Hence, the correct answer is option: 'A'.