If sin-1 x + sin-1 y = \(\frac{2\pi}{3}\) then (cos-1 x + cos-1 y) =

1.	If sin-1 x + sin-1 y = \(\frac{2\pi}{3}\) then (cos-1 x + cos-1 y) = ?A. \(\frac{\pi}{6}\)B. \(\frac{\pi}{3}\) C. \(\pi\)D. \(\frac{2\pi}{3}\)
Answer» Correct Answer is (B) \(\frac{\pi}{3}\) Given: sin^-1 x + sin^-1 y = \(\frac{2\pi}{3}\) Since we know that sin^-1x + cos^-1 x = \(\frac{\pi}{2}\) ⇒ cos^-1 x = \(\frac{\pi}{2}\)- sin^-1x Similarly, cos^-1y = \(\frac{\pi}{2}\)- sin^-1 y Now consider cos^-1 x + cos^-1 y = \(\frac{\pi}{2}\)- sin^-1 x + \(\frac{\pi}{2}\) - sin^-1 y = \(\frac{2\pi}{2}\) - [sin^-1 x + sin^-1 y] = π - \(\frac{2\pi}{2}\) = \(\frac{\pi}{3}\)

If sin-1 x + sin-1 y = \(\frac{2\pi}{3}\) then (cos-1 x + cos-1 y) = ?A. \(\frac{\pi}{6}\)B. \(\frac{\pi}{3}\) C. \(\pi\)D. \(\frac{2\pi}{3}\)

Answer»

Correct Answer is (B) \(\frac{\pi}{3}\)

Given: sin^-1 x + sin^-1 y = \(\frac{2\pi}{3}\)

Since we know that sin^-1x + cos^-1 x = \(\frac{\pi}{2}\)

⇒ cos^-1 x = \(\frac{\pi}{2}\)- sin^-1x

Similarly, cos^-1y = \(\frac{\pi}{2}\)- sin^-1 y

Now consider cos^-1 x + cos^-1 y = \(\frac{\pi}{2}\)- sin^-1 x + \(\frac{\pi}{2}\) - sin^-1 y

= \(\frac{2\pi}{2}\) - [sin^-1 x + sin^-1 y]

= π - \(\frac{2\pi}{2}\)

= \(\frac{\pi}{3}\)