
How do you solve the following equation \[{{\tan }^{-1}}\dfrac{1-x}{1+x}=\dfrac{1}{2}{{\tan }^{-1}}x\] and \[x>0\]?
Answer
543.3k+ views
Hint: 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 \[{{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\dfrac{a-b}{a+b}\]. 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,
\[{{\tan }^{-1}}\dfrac{1-x}{1+x}=\dfrac{1}{2}{{\tan }^{-1}}x\]
Using the trigonometric identity i.e. \[{{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\dfrac{a-b}{a+b}\]
Applying the identity in the above given trigonometric equation, we obtained
\[{{\tan }^{-1}}1-{{\tan }^{-1}}x=\dfrac{1}{2}{{\tan }^{-1}}x\]
Adding \[{{\tan }^{-1}}x\] to both the sides of the equation, we get
\[{{\tan }^{-1}}1-{{\tan }^{-1}}x+{{\tan }^{-1}}x=\dfrac{1}{2}{{\tan }^{-1}}x+{{\tan }^{-1}}x\]
Simplifying the terms in the above equation, we get
\[{{\tan }^{-1}}1=\dfrac{1}{2}{{\tan }^{-1}}x+{{\tan }^{-1}}x\]
Solving the RHS of the above equation, we obtained
\[{{\tan }^{-1}}1=\dfrac{3}{2}{{\tan }^{-1}}x\]
As we know that the value of\[{{\tan }^{-1}}=\dfrac{\pi }{4}\].
Therefore,
\[\dfrac{\pi }{4}=\dfrac{3}{2}{{\tan }^{-1}}x\]
Simplifying the above equation, we obtained
\[\dfrac{\pi }{6}={{\tan }^{-1}}x\]
Removing the inverse of tan, we obtained
\[\tan \dfrac{\pi }{6}=x\]
Using the trigonometric ratios table,
We know that the value of \[\tan \dfrac{\pi }{6}\]is \[\dfrac{1}{\sqrt{3}}\].
Therefore,
\[\Rightarrow x=\dfrac{1}{\sqrt{3}}\]
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.
Complete step by step solution:
We have given that,
\[{{\tan }^{-1}}\dfrac{1-x}{1+x}=\dfrac{1}{2}{{\tan }^{-1}}x\]
Using the trigonometric identity i.e. \[{{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\dfrac{a-b}{a+b}\]
Applying the identity in the above given trigonometric equation, we obtained
\[{{\tan }^{-1}}1-{{\tan }^{-1}}x=\dfrac{1}{2}{{\tan }^{-1}}x\]
Adding \[{{\tan }^{-1}}x\] to both the sides of the equation, we get
\[{{\tan }^{-1}}1-{{\tan }^{-1}}x+{{\tan }^{-1}}x=\dfrac{1}{2}{{\tan }^{-1}}x+{{\tan }^{-1}}x\]
Simplifying the terms in the above equation, we get
\[{{\tan }^{-1}}1=\dfrac{1}{2}{{\tan }^{-1}}x+{{\tan }^{-1}}x\]
Solving the RHS of the above equation, we obtained
\[{{\tan }^{-1}}1=\dfrac{3}{2}{{\tan }^{-1}}x\]
As we know that the value of\[{{\tan }^{-1}}=\dfrac{\pi }{4}\].
Therefore,
\[\dfrac{\pi }{4}=\dfrac{3}{2}{{\tan }^{-1}}x\]
Simplifying the above equation, we obtained
\[\dfrac{\pi }{6}={{\tan }^{-1}}x\]
Removing the inverse of tan, we obtained
\[\tan \dfrac{\pi }{6}=x\]
Using the trigonometric ratios table,
We know that the value of \[\tan \dfrac{\pi }{6}\]is \[\dfrac{1}{\sqrt{3}}\].
Therefore,
\[\Rightarrow x=\dfrac{1}{\sqrt{3}}\]
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.
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