Question

If \( \sin^{-1}x+\sin^{-1}y=\dfrac{2\pi}{3} \), then \( \cos^{-1}x+\cos^{-1}y \) is equal to

• A. \( -\dfrac{\pi}{2} \)
• B. \( \dfrac{\pi}{2} \)
• C. \( \pi \)
• D. \( \dfrac{2\pi}{3} \)
• E. \( \dfrac{\pi}{3} \)

Hint:

Remember the identity \( \sin^{-1}t+\cos^{-1}t=\frac{\pi}{2} \). It directly converts sums of inverse sine into sums of inverse cosine and vice versa.

Answer 1

Answer: (E)

Step 1: Recall the relationship between \( \sin^{-1}t \) and \( \cos^{-1}t \).
For every \( t \in [-1,1] \), we have \[ \sin^{-1}t+\cos^{-1}t=\frac{\pi}{2} \] This is a very important standard identity.

Step 2: Apply the identity separately to \( x \) and \( y \).
Thus, \[ \cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x \] and \[ \cos^{-1}y=\frac{\pi}{2}-\sin^{-1}y \]

Step 3: Add the two equations.
Adding both, we get \[ \cos^{-1}x+\cos^{-1}y = \left(\frac{\pi}{2}-\sin^{-1}x\right) + \left(\frac{\pi}{2}-\sin^{-1}y\right) \] \[ =\pi-\left(\sin^{-1}x+\sin^{-1}y\right) \]

Step 4: Substitute the given value.
We are given \[ \sin^{-1}x+\sin^{-1}y=\frac{2\pi}{3} \] Hence, \[ \cos^{-1}x+\cos^{-1}y = \pi-\frac{2\pi}{3} \]

Step 5: Simplify the expression.
\[ \pi-\frac{2\pi}{3} = \frac{3\pi}{3}-\frac{2\pi}{3} = \frac{\pi}{3} \]

Step 6: Check consistency with principal ranges.
Since both \( \cos^{-1}x \) and \( \cos^{-1}y \) lie in \[ [0,\pi] \] their sum must be non-negative.
The value \[ \frac{\pi}{3} \] is perfectly valid, while negative options are impossible.

Step 7: Final conclusion.
Therefore, \[ \boxed{\cos^{-1}x+\cos^{-1}y=\frac{\pi}{3}} \] Hence, the correct option is \[ \boxed{(5)\ \frac{\pi}{3}} \]