Question

Solve for x where x lies between 0 and 1, \[{{\tan }^{-1}}\left( \dfrac{1+x}{1-x} \right)=\dfrac{\pi }{4}+{{\tan }^{-1}}x\]

Answer 1

Hint: We will first rearrange the given expression and then begin with the left hand side of the given expression and then we will first apply the formula \[{{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)\] on the two terms in the left hand side.

Complete step-by-step answer:
Rearranging the given expression first and then the Left hand side of the given expression is \[{{\tan }^{-1}}\left( \dfrac{1+x}{1-x} \right)-{{\tan }^{-1}}x.......(1)\]
Now we know the formula that \[{{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)\]. So applying this formula to the terms in equation (1) we get,
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{1+x}{1-x}-x}{1+\dfrac{1+x}{1-x}\times x} \right).......(2)\]
Now taking the LCM in the numerator and simplifying all the terms by multiplying in equation (2) we get,
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{1+x-x(1-x)}{1-x}}{\dfrac{1-x+x(1+x)}{1-x}} \right).......(3)\]
Now again simplifying the terms in equation (3) by cancelling similar terms we get,\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{1+x-x(1-x)}{1-x+x(1+x)} \right).......(4)\]
Now opening the brackets and multiplying the terms in equation (4) we get,
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{1+x-x+{{x}^{2}}}{1-x+x+{{x}^{2}}} \right).......(5)\]
Again cancelling similar terms in both numerator and denominator in equation (5) we get,
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{1+{{x}^{2}}}{1+{{x}^{2}}} \right).......(6)\]
Now we see that the both the numerator and denominator is same and hence cancelling them in equation (6) we get,
\[\Rightarrow {{\tan }^{-1}}\left( 1 \right).......(7)\]
Now we know that \[\tan \dfrac{\pi }{4}\] is 1. So substituting this in place of 1 in equation (7) we get,
\[\Rightarrow {{\tan }^{-1}}\left( \tan \dfrac{\pi }{4} \right)=\dfrac{\pi }{4}.....(8)\]
Now from equation (8) we can see that the value \[\dfrac{\pi }{4}\] in the left hand side is equal to \[\dfrac{\pi }{4}\]. Hence it is true for all real values of x between 0 and 1.

Note: Remembering the properties and formulas of inverse trigonometric equations is the key here. We in a hurry can make a mistake in applying the formula in equation (2) so we need to be very careful while doing this step. Also we need to understand that when we get the left hand side equal to the right hand side it means this satisfies all real values of x.