Question

Solve for $ x,\,\,{{\tan }^{-1}}(1/x)=\pi +{{\tan }^{-1}}x,\,\,0 < x < 1 $

• A. $not\, defined$
• B. $ \sqrt{3} $
• C. $ \pm \,1 $
• D. None of the above

Answer 1

The Correct Option is A

We have, $ {{\tan }^{-1}}\left( \frac{1}{x} \right)=\pi +{{\tan }^{-1}}x,\,0 < x < 1 $
$ \Rightarrow $ $ {{\tan }^{-1}}\left( \frac{1}{x} \right)-{{\tan }^{-1}}x=\pi $
$ \Rightarrow $ $ {{\tan }^{-1}}\left( \frac{\frac{1}{x}-x}{1+\frac{1}{x}.x} \right)=\pi $
$ \Rightarrow $ $ {{\tan }^{-1}}\left( \frac{1-{{x}^{2}}}{x(1+1)} \right)=\pi $
$ \Rightarrow $ $ \frac{1-{{x}^{2}}}{2x}=\tan \pi $
$ \Rightarrow $ $ \frac{1-{{x}^{2}}}{2x}=0 $
$ \Rightarrow $ $ 1-{{x}^{2}}=0\,\Rightarrow {{x}^{2}}=1\,\Rightarrow x=\pm 1 $ but given, $ 0 < x < 1 $

Answer 2

Ans. The tangent function has an inverse Tan-1 (x)that works in reverse. It is used to determine the angle when the perpendicular and base of a triangle are given. The Derivative of tan-1x is 1/(1 + x2) is used in solving numerous questions. The function can be integrated using integration by parts. With this formula, we can find the value of an angle when the value of tangent is given.

Considering the trigonometric identity-:

When we put a = tan^-1x,b =  tan^-1y, we get 

tan^-1x + tan^-1y = tan^-1(x + y)/(1 - xy)

tan^-1x - tan^-1y = tan^-1(x - y)/(1 + xy)

Inverse Tangent Function in Derivation

Inverse tangent is used to solve the problems based on derivation. The derivative of tan^-1x is denoted by d.tan^-1x /dx. The derivative of tan^-1x comes out to be 1/1+x². The derivative of tan^-1x is useful for defining the graph plotted between π/2 and –π/2. 

Inverse Tangent Function in Integration

Inverse Tangent Function is used in solving problems based on integration. There is no direct way to integrate tan^-1x. However, tan^-1x can be integrated using integration by parts. The integration of tan^-1x can be calculated as follows-:

So, now we need to integrate tan^-1x.

We know that d(tan^-1x)/dx = 1/(1 + x²) 

We will use the formula ∫uv dx = u ∫vdx - ∫[du/dx. ∫vdx] dx