Solving for the tangent of an inverse secant function
Hello, I am trying to evaluate this expression but haven't figured out how...
$$\tan(\sec^{-1}(x))$$
In Wolfram Alpha it evaluates to the following :
$$\sqrt{1 - \frac 1 {x^2}} x $$
How does one arrive at this ? I am using this in the context of trig substitution.
On
\begin{align} & x = \sec \theta = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{\text{hypotenuse}} 1 \\[10pt] & \sec^{-1} x = \theta = \sec^{-1} \left( \frac{\text{hypotenuse}}{\text{adjacent}} \right) = \sec^{-1} \left( \frac{\text{hypotenuse}} 1 \right) \\[10pt] & \tan\sec^{-1} x = \tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{\text{opposite}} 1 = \frac{\sqrt{\text{hypotenuse}^2 - \text{adjacent}^2}} 1 = \sqrt{x^2-1} \end{align} The above works in the first quadrant, where the tangent is positive.
Where the tangent is negative, use the identity $\sqrt{x^2-1} = \sqrt{1-\dfrac 1 {x^2}}\cdot \sqrt{x^2} = \sqrt{1 - \dfrac 1 {x^2}} \cdot |x|.$
\begin{align} \tan\sec^{-1}x &= \dfrac{\sin\sec^{-1}x}{\cos\sec^{-1}x} \\ &= \dfrac{\sqrt{1-\cos^2\sec^{-1}x}}{\cos\sec^{-1}x} \\ &= \sec\sec^{-1}x\sqrt{1-\dfrac{1}{\sec^2\sec^{-1}x}} \\ &= x\sqrt{1-\dfrac{1}{x^2}} \end{align}