Question

If $\sin^{-1}x+\cos^{-1}y=\frac{3\pi}{10}$, then the value of $\cos^{-1}x+\sin^{-1}y$ is

• A. $\frac{\pi}{10}$
• B. $\frac{7\pi}{10}$
• C. $\frac{9\pi}{10}$
• D. $\frac{3\pi}{10}$

Hint:

Logic Tip: An even faster method is to simply set up a system of equations. Let $E_1 = \sin^{-1}x + \cos^{-1}y$ and $E_2 = \cos^{-1}x + \sin^{-1}y$. If you add equations $E_1$ and $E_2$, you get $(\sin^{-1}x+\cos^{-1}x) + (\cos^{-1}y+\sin^{-1}y) = \frac{\pi}{2} + \frac{\pi}{2} = \pi$. Therefore, $E_2 = \pi - E_1$.

Answer 1

The Correct Option is B

Concept:
The fundamental complementary angle identities for inverse trigonometric functions are: $$\sin^{-1}z + \cos^{-1}z = \frac{\pi}{2} \quad \text{for } z \in [-1, 1]$$ This means we can express $\sin^{-1}z$ as $\frac{\pi}{2} - \cos^{-1}z$ and vice versa.
Step 1: Write down the given equation and the identities.
Given equation: $$\sin^{-1}x + \cos^{-1}y = \frac{3\pi}{10}$$ We know the identities: $$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \implies \sin^{-1}x = \frac{\pi}{2} - \cos^{-1}x$$ $$\sin^{-1}y + \cos^{-1}y = \frac{\pi}{2} \implies \cos^{-1}y = \frac{\pi}{2} - \sin^{-1}y$$
Step 2: Substitute the identities into the given equation.
Substitute the expressions for $\sin^{-1}x$ and $\cos^{-1}y$ into the given equation: $$\left(\frac{\pi}{2} - \cos^{-1}x\right) + \left(\frac{\pi}{2} - \sin^{-1}y\right) = \frac{3\pi}{10}$$
Step 3: Rearrange to solve for the required expression.
Group the $\frac{\pi}{2}$ terms and the inverse trigonometric terms: $$\left(\frac{\pi}{2} + \frac{\pi}{2}\right) - (\cos^{-1}x + \sin^{-1}y) = \frac{3\pi}{10}$$ $$\pi - (\cos^{-1}x + \sin^{-1}y) = \frac{3\pi}{10}$$ Now, isolate the required expression $(\cos^{-1}x + \sin^{-1}y)$: $$\cos^{-1}x + \sin^{-1}y = \pi - \frac{3\pi}{10}$$ $$\cos^{-1}x + \sin^{-1}y = \frac{10\pi - 3\pi}{10}$$ $$\cos^{-1}x + \sin^{-1}y = \frac{7\pi}{10}$$