find the value of 2 tan⁻¹(1/3) + tan⁻¹(1/7)

• tan⁻¹(49/29)
• π/2
• 0
• π/4

To solve the expression \(2 \tan^{-1}(1/3) + \tan^{-1}(1/7) + \tan^{-1}(49/29)\), we can use the properties of inverse tangent functions.

Step 1: Simplifying \(2 \tan^{-1}(1/3)\)

Using the double angle formula for tangent, we have:

tan(2θ) = 2tan(θ) / (1 - tan²(θ))

Let \(θ = \tan^{-1}(1/3)\), then:

tan(2θ) = \frac{2(1/3)}{1 - (1/3)²} = \frac{2/3}{1 - 1/9} = \frac{2/3}{8/9} = \frac{6}{8} = \frac{3}{4}

Step 2: Adding \(\tan^{-1}(1/7)\)

Now we need to find:

tan(2θ + \tan^{-1}(1/7))

Using the formula:

tan(A + B) = \frac{tan(A) + tan(B)}{1 - tan(A)tan(B)}

Let \(A = 2θ\) and \(B = \tan^{-1}(1/7)\), then:

tan(A + B) = \frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4} \cdot \frac{1}{7}} = \frac{\frac{21}{28} + \frac{4}{28}}{1 - \frac{3}{28}} = \frac{\frac{25}{28}}{\frac{25}{28}} = 1

Step 3: Final Addition with \(\tan^{-1}(49/29)\)

Now we add \(\tan^{-1}(49/29)\):

Since \(tan(A + B) = 1\), we have:

A + B = \tan^{-1}(1) = \frac{π}{4}

Next, we find:

tan\left(\frac{π}{4} + \tan^{-1}\left(\frac{49}{29}\right)\right)

Using the addition formula again:

tan\left(\frac{π}{4} + \tan^{-1}\left(\frac{49}{29}\right)\right) = \frac{1 + \frac{49}{29}}{1 - 1 \cdot \frac{49}{29}} = \frac{\frac{29 + 49}{29}}{\frac{29 - 49}{29}} = \frac{78/29}{-20/29} = -\frac{78}{20} = -\frac{39}{10}

Final Result

Since the tangent of the resulting angle is negative, we conclude that:

2 tan⁻¹(1/3) + tan⁻¹(1/7) + tan⁻¹(49/29) = π/2

Thus, the value is:

π/2