Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

For (P): Let ${\sin }^{-1}x=A$ and ${\sin }^{-1}y=B$. The given equations become $A+B=2\pi /3$ and $(\pi /2-A)-(\pi /2-B)=\pi /3⟹B-A=\pi /3$. Solving these gives $A=\pi /6,B=\pi /2$. Thus $x=\sin (\pi /6)=1/2$ and $y=\sin (\pi /2)=1$, so $x+y=3/2$. For (Q): Using ${\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)$, the equation becomes $\frac{2x}{x^2-2}=\frac{1}{2}$, leading to $x^2-4x-2=0$. The solutions are $x_{1,2}=2\pm \sqrt{6}$, both valid. Then $x_1^2+x_2^2=(2-\sqrt{6}{)}^2+(2+\sqrt{6}{)}^2=(4+6-4\sqrt{6})+(4+6+4\sqrt{6})=20$. For (R): The equation ${\sin }^{-1}(\sin x)={\cos }^{-1}(\cos x)$ holds when $x\in [0,\pi ]$ and $x\in [2\pi ,3\pi ]$. In $[0,\pi ]$, integral values are $0,1,2,3$ (4 values). In $[2\pi ,3\pi ]$, integral values are $7,8,9$ (3 values). Total is $4+3=7$. For (S): $P={\sum }_{n=1}^{\infty }{\cot }^{-1}(n^2+n+1)={\sum }_{n=1}^{\infty }{\tan }^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right)={\sum }_{n=1}^{\infty }({\tan }^{-1}(n+1)-{\tan }^{-1}(n))$. This is a telescoping sum, which evaluates to ${\lim }_{N\to \infty }({\tan }^{-1}(N+1)-{\tan }^{-1}(1))=\pi /2-\pi /4=\pi /4$. Therefore, $P/\pi =(\pi /4)/\pi =1/4$.