Question

solve the equation for exact solutions.

$$\cos^{-1}x - \tan^{-1}x = \frac{\pi}{2}$$

solve the equation for exact solutions. select the correct choice below and, if necessary, fill in

a. the solution set is \\( \\{ \\} \\).
(simplify your answer, including any radicals. use integers or fractions for any number

b. the solution is the empty set.

Explanation:

Step1: Use the property of inverse - trigonometric functions

We know that \(\cos^{-1}x+\sin^{-1}x=\frac{\pi}{2}\) for \(x\in[- 1,1]\). So, the given equation \(\cos^{-1}x-\tan^{-1}x=\frac{\pi}{2}\) can be rewritten as \(\sin^{-1}x+\cos^{-1}x-(\sin^{-1}x+\tan^{-1}x)=0\). Since \(\cos^{-1}x+\sin^{-1}x=\frac{\pi}{2}\), we have \(\sin^{-1}x+\tan^{-1}x = 0\).

Step2: Use the definitions of inverse - trigonometric functions

Let \(y = \sin^{-1}x\), then \(x=\sin y\) and \(\tan^{-1}x=-y\). So \(x =-\tan y\). Since \(x = \sin y\) and \(x=-\tan y=-\frac{\sin y}{\cos y}\).
If \(x
eq0\), we can cancel \(\sin y\) (because if \(\sin y = 0\), then \(x = 0\) and \(\sin^{-1}(0)+\tan^{-1}(0)=0 + 0=0\)). We get \(1=-\frac{1}{\cos y}\), so \(\cos y=-1\). Then \(y=\pi+2k\pi,k\in\mathbb{Z}\), but the range of \(y = \sin^{-1}x\) is \([-\frac{\pi}{2},\frac{\pi}{2}]\).
If \(x = 0\), \(\sin^{-1}(0)+\tan^{-1}(0)=0+0 = 0\).

Answer:

A. The solution set is \(\{0\}\)