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$\quad$ Consider the following
I. $\quad \operatorname{cosec}^{-1}\left(-\frac{2}{\sqrt{3}}\right)=-\frac{\pi}{3}$
II. $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi}{6}$
Which of the above is/are correct?
Options:
I. $\quad \operatorname{cosec}^{-1}\left(-\frac{2}{\sqrt{3}}\right)=-\frac{\pi}{3}$
II. $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi}{6}$
Which of the above is/are correct?
Solution:
2029 Upvotes
Verified Answer
The correct answer is:
Both I and II
Consider (I):
$$
\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)=\frac{-\pi}{3}
$$
Let $\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)=\theta$
$$
\begin{array}{l}
\Rightarrow \frac{-2}{\sqrt{3}}=\operatorname{cosec} \theta \Rightarrow \frac{-\sqrt{3}}{2}=\frac{1}{\operatorname{cosec} \theta} \\
\Rightarrow \frac{-\sqrt{3}}{2}=\sin \theta \Rightarrow \theta=\frac{-\pi}{3}
\end{array}
$$
Now, consider (II):
$$
\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi}{6}
$$
Let $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\theta$
$\Rightarrow \frac{2}{\sqrt{3}}=\sec \theta \Rightarrow \frac{\sqrt{3}}{2}=\frac{1}{\sec \theta}$
$\Rightarrow \frac{\sqrt{3}}{2}=\cos \theta \quad \Rightarrow \theta=\frac{\pi}{6}$
Hence, both statements I and II are correct.
$$
\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)=\frac{-\pi}{3}
$$
Let $\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)=\theta$
$$
\begin{array}{l}
\Rightarrow \frac{-2}{\sqrt{3}}=\operatorname{cosec} \theta \Rightarrow \frac{-\sqrt{3}}{2}=\frac{1}{\operatorname{cosec} \theta} \\
\Rightarrow \frac{-\sqrt{3}}{2}=\sin \theta \Rightarrow \theta=\frac{-\pi}{3}
\end{array}
$$
Now, consider (II):
$$
\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi}{6}
$$
Let $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\theta$
$\Rightarrow \frac{2}{\sqrt{3}}=\sec \theta \Rightarrow \frac{\sqrt{3}}{2}=\frac{1}{\sec \theta}$
$\Rightarrow \frac{\sqrt{3}}{2}=\cos \theta \quad \Rightarrow \theta=\frac{\pi}{6}$
Hence, both statements I and II are correct.
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