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Question: \( \tan \left( {{{\cos }^{ - 1}}x} \right) \) is equal to \( A.\,\,\dfrac{x}{{1 + {x^2}}} \...

tan⁡(cos⁡−1x)\tan \left( {{{\cos }^{ - 1}}x} \right) is equal to
A.  x1+x2 B.  1+x2x C.  1−x2x D.  1−2x  A.\,\,\dfrac{x}{{1 + {x^2}}} \\\ B.\,\,\dfrac{{\sqrt {1 + {x^2}} }}{x} \\\ C.\,\,\dfrac{{\sqrt {1 - {x^2}} }}{x} \\\ D.\,\,\sqrt {1 - 2x} \\\

Explanation

Solution

Hint : For this type of trigonometric equation having angle in terms of inverse trigonometric. We first let the inverse function as A then using trigonometric identities we used it to find the value of other trigonometric functions and hence the solution of the given problem.
Formulas used: sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1

Complete step-by-step answer :
Given equation is
tan⁡(cos⁡−1x)\tan \left( {{{\cos }^{ - 1}}x} \right)
As we see that the angle of tan is cos⁡−1x{\cos ^{ - 1}}x and we know that to find its value we have to simplify the angle.
For this we first let cos⁡−1x=A{\cos ^{ - 1}}x = A
Or we can write it as:
cos⁡A=x\cos A = x
Also, from trigonometric identity we have
sin⁡2A+cos⁡2A=1 or  we  can  write  above  equation  as sin⁡2A=1−cos⁡2A  {\sin ^2}A + {\cos ^2}A = 1 \\\ or\,\,we\,\,can\,\,write\,\,above\,\,equation\,\,as \\\ {\sin ^2}A = 1 - {\cos ^2}A \\\
Substituting value of cos⁡A=x\cos A = x in above equation. We have
sin⁡2A=1−(x)2 or sin⁡2A=1−x2 ⇒sin⁡A=1−x2  {\sin ^2}A = 1 - {\left( x \right)^2} \\\ or \\\ {\sin ^2}A = 1 - {x^2} \\\ \Rightarrow \sin A = \sqrt {1 - {x^2}} \\\
Also, from above we have

tan⁡(cos⁡−1x)=tan⁡(A) ⇒tan⁡(cos⁡−1x)=sin⁡Acos⁡A   \tan \left( {{{\cos }^{ - 1}}x} \right) = \tan \left( A \right) \\\ \Rightarrow \tan \left( {{{\cos }^{ - 1}}x} \right) = \dfrac{{\sin A}}{{\cos A}} \;

Now, substituting the value of sin⁡A  and  cos⁡A\sin A\,\,and\,\,\cos A calculated above in the formed equation.
We have,
tan⁡(cos⁡−1x)=1−x2x\tan \left( {{{\cos }^{ - 1}}x} \right) = \dfrac{{\sqrt {1 - {x^2}} }}{x}
Therefore, from above we see that required value of tan⁡(cos⁡−1x)\tan \left( {{{\cos }^{ - 1}}x} \right) is 1−x2x\dfrac{{\sqrt {1 - {x^2}} }}{x} .
So, the correct answer is “Option C”.

Note : In this type of problem we first take inverse function as some other variable and then finding trigonometric function in term of variable taken and then using trigonometric identities we will find other trigonometric function in term of assumed or taken variable and then substituting values calculated in given to find required value or required solution of given problem.