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Let $(x, y)$ be such that $\sin ^{-1}(a x)+\cos ^{-1}(y)+\cos ^{-1}(b x y)=\frac{\pi}{2}$. Match the statements in Column I with the values in Column II.
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Verified Answer
The correct answer is:
A-p; B-q; C-p; D-s
A-p; B-q; C-p; D-s
(A) If $a=1$ and $b=0$, then $\sin ^{-1} x+\cos ^{-1} y=0$
$$
\Rightarrow \quad \sin ^{-1} x=-\cos ^{-1} y \Rightarrow x^2+y^2=1
$$
(B) If $a=1$ and $b=1$, then
$$
\begin{array}{rlrl}
& & \sin ^{-1} x+\cos ^{-1} y+\cos ^{-1} x y & =\frac{\pi}{2} \\
\Rightarrow & & \cos ^{-1} x-\cos ^{-1} y & =\cos ^{-1} x y \\
\Rightarrow & x y+\sqrt{1-x^2} \sqrt{1-y^2} & =x y \quad \text { [taking sine on both the sides] }
\end{array}
$$
(C) If $a=1$ and $b=2$, then
$$
\begin{aligned}
& & \sin ^{-1} x+\cos ^{-1} y+\cos ^{-1}(2 x y) & =\frac{\pi}{2} \\
\Rightarrow & & \cos ^{-1} x-\cos ^{-1} y & =\cos ^{-1}(2 x y) \\
\Rightarrow & & x y+\sqrt{1-x^2} \sqrt{1-y^2} & =2 x y \\
\Rightarrow & & x^2+y^2 & =1
\end{aligned}
$$
[on squaring]
(D) If $a=2$ and $b=2$, then
$$
\begin{aligned}
& & \sin ^{-1}(2 x)+\cos ^{-1}(y)+\cos ^{-1}(2 x y) & =\frac{\pi}{2} \\
\Rightarrow & & 2 x y+\sqrt{1+4 x^2} \sqrt{1-y^2} & =2 x y \\
\Rightarrow & & \left(4 x^2-1\right)\left(y^2-1\right) & =0 .
\end{aligned}
$$
$$
\Rightarrow \quad \sin ^{-1} x=-\cos ^{-1} y \Rightarrow x^2+y^2=1
$$
(B) If $a=1$ and $b=1$, then
$$
\begin{array}{rlrl}
& & \sin ^{-1} x+\cos ^{-1} y+\cos ^{-1} x y & =\frac{\pi}{2} \\
\Rightarrow & & \cos ^{-1} x-\cos ^{-1} y & =\cos ^{-1} x y \\
\Rightarrow & x y+\sqrt{1-x^2} \sqrt{1-y^2} & =x y \quad \text { [taking sine on both the sides] }
\end{array}
$$
(C) If $a=1$ and $b=2$, then
$$
\begin{aligned}
& & \sin ^{-1} x+\cos ^{-1} y+\cos ^{-1}(2 x y) & =\frac{\pi}{2} \\
\Rightarrow & & \cos ^{-1} x-\cos ^{-1} y & =\cos ^{-1}(2 x y) \\
\Rightarrow & & x y+\sqrt{1-x^2} \sqrt{1-y^2} & =2 x y \\
\Rightarrow & & x^2+y^2 & =1
\end{aligned}
$$
[on squaring]
(D) If $a=2$ and $b=2$, then
$$
\begin{aligned}
& & \sin ^{-1}(2 x)+\cos ^{-1}(y)+\cos ^{-1}(2 x y) & =\frac{\pi}{2} \\
\Rightarrow & & 2 x y+\sqrt{1+4 x^2} \sqrt{1-y^2} & =2 x y \\
\Rightarrow & & \left(4 x^2-1\right)\left(y^2-1\right) & =0 .
\end{aligned}
$$
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