Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

(A) If $a=1$ and $b=0$, then ${\sin }^{-1}x+{\cos }^{-1}y=0$ $$\Rightarrow {\sin }^{-1}x=-{\cos }^{-1}y\Rightarrow x^2+y^2=1$$ (B) If $a=1$ and $b=1$, then $$\begin{array}{rlrl}& & {\sin }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}xy& =\frac{\pi }{2}\\ \Rightarrow & & {\cos }^{-1}x-{\cos }^{-1}y& ={\cos }^{-1}xy\\ \Rightarrow & xy+\sqrt{1-x^2}\sqrt{1-y^2}& =xy\text{ [taking sine on both the sides] }& \end{array}$$ (C) If $a=1$ and $b=2$, then $$\begin{array}{rlrl}& & {\sin }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}(2xy)& =\frac{\pi }{2}\\ \Rightarrow & & {\cos }^{-1}x-{\cos }^{-1}y& ={\cos }^{-1}(2xy)\\ \Rightarrow & & xy+\sqrt{1-x^2}\sqrt{1-y^2}& =2xy\\ \Rightarrow & & x^2+y^2& =1\end{array}$$ [on squaring] (D) If $a=2$ and $b=2$, then $$\begin{array}{rlrl}& & {\sin }^{-1}(2x)+{\cos }^{-1}(y)+{\cos }^{-1}(2xy)& =\frac{\pi }{2}\\ \Rightarrow & & 2xy+\sqrt{1+4x^2}\sqrt{1-y^2}& =2xy\\ \Rightarrow & & \left(4x^2-1\right)\left(y^2-1\right)& =0.\end{array}$$