Question

If $\cos^{-1}x + \cos^{-1}y = 2\pi$, then the value of $\sin^{-1}x + \sin^{-1}y$ is equal to

• A. $0$
• B. $\pi$
• C. $-\pi$
• D. None of these

Hint:

Range of $\cos^{-1}x$ is $[0,\pi]$, range of $\sin^{-1}x$ is $[-\pi/2, \pi/2]$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
$\cos^{-1}x \in [0,\pi]$, so $\cos^{-1}x + \cos^{-1}y = 2\pi$ implies both are $\pi$.
Step 2: Detailed Explanation:
The range of $\cos^{-1}x$ is $[0,\pi]$. The maximum sum is $2\pi$, which occurs only when $\cos^{-1}x = \pi$ and $\cos^{-1}y = \pi$, i.e., $x = -1$ and $y = -1$.
Then $\sin^{-1}(-1) + \sin^{-1}(-1) = -\frac{\pi}{2} - \frac{\pi}{2} = -\pi$. But that's not $0$. However, if $x$ and $y$ are such that $\cos^{-1}x = \pi$, then $x=-1$, similarly $y=-1$. Then $\sin^{-1}x = -\pi/2$, sum = $-\pi$.
So answer should be $-\pi$. But option (C) is $-\pi$.
Let's check: If $\cos^{-1}x = 2\pi - \cos^{-1}y$, the maximum is $2\pi$, so both must be $\pi$. So $x=y=-1$. Then $\sin^{-1}(-1) = -\pi/2$, sum = $-\pi$. So (C) is correct.
But (A) is $0$. Which is right? Let's verify: $\cos^{-1}(-1) = \pi$, so $\pi + \pi = 2\pi$, yes. $\sin^{-1}(-1) = -\pi/2$, sum = $-\pi$. So answer is $-\pi$.
Step 3: Final Answer:
$\sin^{-1}x + \sin^{-1}y = -\pi$.