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Question

Mathematics Question on Inverse Trigonometric Functions

What is tan⁡(cos⁡−1 x)\tan (\cos^{-1} \, x) equal to ?

A

1−x2x\frac{\sqrt{1-x^{2}}}{x}

B

x1+x2\frac{x}{1+x^{2}}

C

1+x2x\frac{\sqrt{1 + x^{2}}}{x}

D

1−x2\sqrt{ 1 - x^2}

Answer

1−x2x\frac{\sqrt{1-x^{2}}}{x}

Explanation

Solution

Let cos⁡−1x=θ\cos^{-1} x = \theta ⇒cos⁡θ=x⇒sin⁡θ=1−x2\Rightarrow \cos\theta = x \Rightarrow \sin \theta = \sqrt{1-x^{2}} ⇒tan⁡θ=1−x2x \Rightarrow \tan \theta = \frac{\sqrt{1 - x^{2}}}{x} and θ=cos⁡−1x \theta = \cos^{-1} x This can be represented by a triangle with hypotenuse = 1 and sides x and 1−x2 \sqrt{1 -x^{2}} . ⇒tan⁡(cos⁡−1x)=1−x2x\Rightarrow \tan\left(\cos^{-1} x \right) = \frac{\sqrt{1-x^{2}}}{x}