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Question: How do you simplify \[\tan \left( {{{\cos }^{ - 1}}x} \right)\]?...

How do you simplify tan⁡(cos⁡−1x)\tan \left( {{{\cos }^{ - 1}}x} \right)?

Explanation

Solution

To solve this question first we will assume cos⁡−1x=θ\cos^{-1}x= \theta . We use standard identity cos⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1 to get the value of sin⁡θ\sin {\theta}. Then using these values we will calculate the required value.

Complete step by step answer:
The given question is tan⁡(cos⁡−1x)\tan \left( {{{\cos }^{ - 1}}x} \right)
Take an assumption that, cos⁡−1x=θ{\cos ^{ - 1}}x = \theta
Which implies that cos⁡θ=x\cos \theta = x
We all know that cos⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1
So, from here, we can write sin⁡θ=±1−cos⁡2x\sin \theta = \pm \sqrt {1 - {{\cos }^2}x}
⇒sin⁡θ=±1−x2\Rightarrow \sin \theta = \pm \sqrt {1 - {x^2}}
So, the expression becomes tan⁡θ\tan \theta
We all know that tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}
So, we can write tan⁡θ=±1−x2x\tan \theta = \pm \dfrac{{\sqrt {1 - {x^2}} }}{x}
Finally, we can conclude that, tan⁡(cos⁡−1x)=±1−x2x\tan \left( {{{\cos }^{ - 1}}x} \right) = \pm \dfrac{{\sqrt {1 - {x^2}} }}{x}

Note:
Remember the identity cos⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1 which is a standard identity in trigonometry. The result we got consists of both positive and negative values. We have to consider both the values.
sin⁡−1x{\sin ^{ - 1}}x and cos⁡−1x{\cos ^{ - 1}}x are defined only when −1⩽x⩽1 - 1 \leqslant x \leqslant 1.
And also remember the identity sin⁡−1x+cos⁡−1x=π2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}.