Question

\(2\tan^{-1} \frac12+\tan^{-1}\frac 17=\tan^{-1}x\), then the value of \(x\) is:

• A. \(\frac {31}{17}\)
• B. \(\frac {17}{31}\)
• C. \(\frac {4}{3}\)
• D. \(\frac {3}{4}\)

Answer 1

The Correct Option is A

To solve the equation \(2\tan^{-1} \frac{1}{2}+\tan^{-1} \frac{1}{7}=\tan^{-1} x\), we can use the tangent addition formulas. First, let's denote \(a=\tan^{-1} \frac{1}{2}\) and \(b=\tan^{-1} \frac{1}{7}\). Then the equation becomes \(2a+b=\tan^{-1} x\).

Using the tangent double angle identity: \(2\tan^{-1}a=\tan^{-1}\left(\frac{2\tan a}{1-\tan^2 a}\right)\), where \(a=\frac{1}{2}\) implies \(\tan a=\frac{1}{2}\). Hence, \(2\tan^{-1} \frac{1}{2}=\tan^{-1}\left(\frac{2\times\frac{1}{2}}{1-\left(\frac{1}{2}\right)^2}\right)=\tan^{-1}\left(\frac{1}{\frac{3}{4}}\right)=\tan^{-1}\left(\frac{4}{3}\right)\).

Now \(2a+b=\tan^{-1} \frac{4}{3}+\tan^{-1} \frac{1}{7}=\tan^{-1} x\).

Use the tangent addition formula: \(\tan^{-1}u+\tan^{-1}v=\tan^{-1}\left(\frac{u+v}{1-uv}\right)\). Applying it: \(\tan^{-1} \frac{4}{3}+\tan^{-1} \frac{1}{7}=\tan^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\left(\frac{4}{3}\right)\left(\frac{1}{7}\right)}\right)\).

Calculate numerator: \(\frac{4}{3}+\frac{1}{7}=\frac{28}{21}+\frac{3}{21}=\frac{31}{21}\).

Calculate denominator: \(1-\frac{4}{3}\cdot\frac{1}{7}=1-\frac{4}{21}=\frac{17}{21}\).

Therefore, \(\frac{31}{21}\div\frac{17}{21}=\frac{31}{17}\).

So, \(x=\frac{31}{17}\), which confirms that the value of \(x\) is \(\frac{31}{17}\).