Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

The domain for the equation is $x\in [-1,1]$. Let ${\sin }^{-1}(x)=\theta$. Then $x=\sin \theta$ and $\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$. The equation becomes $\theta =(\frac{\pi }{2}-\theta )+{\tan }^{-1}(\sin \theta )$, which simplifies to $2\theta -\frac{\pi }{2}={\tan }^{-1}(\sin \theta )$. Let $f(\theta )=2\theta -\frac{\pi }{2}$ and $g(\theta )={\tan }^{-1}(\sin \theta )$. We need to find the number of solutions for $f(\theta )=g(\theta )$. For $\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$, $f(\theta )$ is a linear function increasing from $-\frac{3\pi }{2}$ to $\frac{\pi }{2}$. $g(\theta )$ is an increasing function from ${\tan }^{-1}(-1)=-\frac{\pi }{4}$ to ${\tan }^{-1}(1)=\frac{\pi }{4}$. At $\theta =0$, $f(0)=-\frac{\pi }{2}$ and $g(0)=0$, so $f(0)<g(0)$. At $\theta =\frac{\pi }{2}$, $f(\frac{\pi }{2})=\frac{\pi }{2}$ and $g(\frac{\pi }{2})=\frac{\pi }{4}$, so $f(\frac{\pi }{2})>g(\frac{\pi }{2})$. Since $f(\theta )$ and $g(\theta )$ are continuous and $f(\theta )$ starts below $g(\theta )$ and ends above it, there must be at least one solution. Now consider derivatives: $f^{\prime }(\theta )=2$ and $g^{\prime }(\theta )=\frac{\cos \theta }{1+{\sin }^2\theta }$. For $\theta \in [0,\frac{\pi }{2}]$, $\cos \theta \ge 0$ and $1+{\sin }^2\theta \ge 1$, so $0\le g^{\prime }(\theta )\le 1$. Since $f^{\prime }(\theta )=2>g^{\prime }(\theta )$ for $\theta \in [0,\frac{\pi }{2})$, if they cross once, they cannot cross again. Similarly for $\theta \in [-\frac{\pi }{2},0]$, $g^{\prime }(\theta )\ge 0$. Thus, there is exactly one solution.