If $0 \leq \theta \leq 2 \pi, 0 \leq \alpha \leq 2 \pi$ and $\sec ^{2018} \theta$…
If $0 \leq \theta \leq 2 \pi, 0 \leq \alpha \leq 2 \pi$ and $\sec ^{2018} \theta$ $+\operatorname{cosec}^{2018} \alpha=2$, then the value of $\cos ^{2020} \theta+\sin ^{2022} \alpha=$
$\frac{3}{2}$
$\frac{1}{2^{2020}}$
1
2
Solution
$\sec ^{2018} \theta+\operatorname{cosec}^{2018} \alpha=2$
It is true for $\theta=0$ and $\alpha=\pi / 2$
$
\begin{aligned}
\therefore \cos ^{2020} \theta+\sin ^{2022} \alpha & =\cos ^{2020} 0+\sin ^{2022} \frac{\pi}{2} \\
& =1+1=2
\end{aligned}
$