Question
If $${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \frac{\pi }{2}$$ and $${\cos ^{ - 1}}x - {\cos ^{ - 1}}y = 0,$$ then values $$x$$ and $$y$$ are respectively
A.
$$\frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }}$$
B.
$$\frac{1}{2}, \frac{1}{2}$$
C.
$$\frac{1}{2}, - \frac{1}{2}$$
D.
$$\frac{1}{{\sqrt 2 }}, \frac{1}{{\sqrt 2 }}$$
Answer :
$$\frac{1}{{\sqrt 2 }}, \frac{1}{{\sqrt 2 }}$$
Solution :
$$\eqalign{
& {\text{Given, }}{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \frac{\pi }{2} ......(1)\cr
& {\text{and }}{\cos ^{ - 1}}x - {\cos ^{ - 1}}y = 0 \cr
& \Rightarrow \left( {\frac{\pi }{2} - {{\sin }^{ - 1}}x} \right) - \left( {\frac{\pi }{2} - {{\sin }^{ - 1}}y} \right) = 0 \cr
& \Rightarrow {\sin ^{ - 1}}y - {\sin ^{ - 1}}x = 0 ......(2) \cr
& \Rightarrow {\sin ^{ - 1}}y = {\sin ^{ - 1}}x \cr} $$
From equations (1) and (2) , we get
$$\eqalign{
& 2{\sin ^{ - 1}}x = \frac{\pi }{2} \cr
& \Rightarrow {\sin ^{ - 1}}x = \frac{\pi }{4} \cr
& \Rightarrow x = \frac{1}{{\sqrt 2 }} \cr} $$
From equation (ii)
$$y = \frac{1}{{\sqrt 2 }}$$