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Question: How do you solve the following equation \[{{\tan }^{-1}}\dfrac{1-x}{1+x}=\dfrac{1}{2}{{\tan }^{-1}}x...

How do you solve the following equation tan⁡−11−x1+x=12tan⁡−1x{{\tan }^{-1}}\dfrac{1-x}{1+x}=\dfrac{1}{2}{{\tan }^{-1}}x and x>0x>0?

Explanation

Solution

In the given question, we have asked to solve the given trigonometric equation for the value of ‘x’. In order to find the value of ‘x’, first we need to simplify the given equation by using the trigonometric identity that is tan⁡−1a−tan⁡−1b=tan⁡−1a−ba+b{{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\dfrac{a-b}{a+b}. Later we simplify the equation further and with the help of trigonometric ratios tables we will put the values of tan function and we will get the value for ‘x’.

Complete step by step solution:
We have given that,
tan⁡−11−x1+x=12tan⁡−1x{{\tan }^{-1}}\dfrac{1-x}{1+x}=\dfrac{1}{2}{{\tan }^{-1}}x
Using the trigonometric identity i.e. tan⁡−1a−tan⁡−1b=tan⁡−1a−ba+b{{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\dfrac{a-b}{a+b}
Applying the identity in the above given trigonometric equation, we obtained
tan⁡−11−tan⁡−1x=12tan⁡−1x{{\tan }^{-1}}1-{{\tan }^{-1}}x=\dfrac{1}{2}{{\tan }^{-1}}x
Adding tan⁡−1x{{\tan }^{-1}}x to both the sides of the equation, we get
tan⁡−11−tan⁡−1x+tan⁡−1x=12tan⁡−1x+tan⁡−1x{{\tan }^{-1}}1-{{\tan }^{-1}}x+{{\tan }^{-1}}x=\dfrac{1}{2}{{\tan }^{-1}}x+{{\tan }^{-1}}x
Simplifying the terms in the above equation, we get
tan⁡−11=12tan⁡−1x+tan⁡−1x{{\tan }^{-1}}1=\dfrac{1}{2}{{\tan }^{-1}}x+{{\tan }^{-1}}x
Solving the RHS of the above equation, we obtained
tan⁡−11=32tan⁡−1x{{\tan }^{-1}}1=\dfrac{3}{2}{{\tan }^{-1}}x
As we know that the value oftan⁡−1=π4{{\tan }^{-1}}=\dfrac{\pi }{4}.
Therefore,
π4=32tan⁡−1x\dfrac{\pi }{4}=\dfrac{3}{2}{{\tan }^{-1}}x
Simplifying the above equation, we obtained
π6=tan⁡−1x\dfrac{\pi }{6}={{\tan }^{-1}}x
Removing the inverse of tan, we obtained
tan⁡π6=x\tan \dfrac{\pi }{6}=x
Using the trigonometric ratios table,
We know that the value of tan⁡π6\tan \dfrac{\pi }{6}is 13\dfrac{1}{\sqrt{3}}.

Therefore,
⇒x=13\Rightarrow x=\dfrac{1}{\sqrt{3}}
Hence, this is the required answer.

Note: In order to solve these types of questions, you should always need to remember the properties of trigonometric and the trigonometric ratios as well. It will make questions easier to solve. It is preferred that while solving these types of questions we should carefully examine the pattern of the given function and then you would apply the formulas according to the pattern observed. As if you directly apply the formula it will create confusion ahead and we will get the wrong answer.