Trigonometric Ratios of Complementary Angles - Practice Questions & MCQ

Complementary Angles
1. $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$, for all $x \in[-1,1]$
2. $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$, for all $x \in R$
3. $\sec ^{-1} x+\operatorname{cosec}^{-1} x=\frac{\pi}{2}$ for all $x \in(-\infty,-1] \cup[1, \infty)$

Proof:
1.

Let $\sin ^{-1} x=\theta$
where $\theta \in[-\pi / 2, \pi / 2]$

$
\begin{array}{ll}
\Rightarrow & -\frac{\pi}{2} \leq-\theta \leq \frac{\pi}{2} \\
\Rightarrow & 0 \leq \frac{\pi}{2}-\theta \leq \pi \\
\Rightarrow & \frac{\pi}{2}-\theta \in[0, \pi]
\end{array}
$
Now, $\sin ^{-1} x=\theta$
Or, $\quad x=\sin \theta$
$\Rightarrow \quad x=\cos \left(\frac{\pi}{2}-\theta\right)$
$\Rightarrow \quad \cos ^{-1} \mathrm{x}=\frac{\pi}{2}-\theta \quad[\because \mathrm{x} \in[-1,1]$ and $(\pi / 2-\theta) \in[0, \pi]]$
$\Rightarrow \quad \theta+\cos ^{-1} \mathrm{x}=\frac{\pi}{2}$
From Eqs.(i) and (ii), we get $\sin ^{-1} x+\cos ^{-1} x=\pi / 2$. Similarly, we get the other results.