Question 40 of 42advanced🔧 ApplyLong Answer5 marks

Solve the following equation:

tan⁡−1[1−x1+x]=12tan⁡−1x,(x>0)\tan^{-1}\left[\frac{1-x}{1+x}\right] = \frac{1}{2}\tan^{-1}x, \quad (x > 0)
Correct Answer

x = 1/3\sqrt{3}

Exercise: Miscellaneous Exercise on Chapter 2 | Q: 12 | (Page 14)
For More Understanding

Explanation

This question tests the application of inverse trigonometric identities. The key identity used is:

tan⁡(π4−θ)=1−tan⁡θ1+tan⁡θ\tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan\theta}{1 + \tan\theta}

By substituting x=tan⁡θx = \tan\theta, we can simplify the left-hand side using this identity. The domain restriction x>0x > 0 ensures θ∈(0,π/2)\theta \in (0, \pi/2), which keeps all inverse tangent values within their principal range.

Solution Steps

  1. Step 1: Let tan⁡−1x=θ\tan^{-1}x = \theta, therefore x=tan⁡θx = \tan\theta. Since x>0x > 0, we have θ∈(0,π/2)\theta \in (0, \pi/2).

  2. Step 2: The given equation becomes:

    tan⁡−1[1−x1+x]=θ2\tan^{-1}\left[\frac{1-x}{1+x}\right] = \frac{\theta}{2}

  3. Step 3: Substitute x=tan⁡θx = \tan\theta in LHS:

    tan⁡−1[1−tan⁡θ1+tan⁡θ]\tan^{-1}\left[\frac{1-\tan\theta}{1+\tan\theta}\right]

  4. Step 4: Using the identity tan⁡(π4−θ)=1−tan⁡θ1+tan⁡θ\tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan\theta}{1 + \tan\theta}, we get:

    tan⁡−1[tan⁡(π4−θ)]\tan^{-1}\left[\tan\left(\frac{\pi}{4} - \theta\right)\right]

  5. Step 5: Since θ∈(0,π/2)\theta \in (0, \pi/2), we have π4−θ∈(−π4,π4)\frac{\pi}{4} - \theta \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right), which lies in the principal range of tan⁡−1\tan^{-1}.

  6. Step 6: Therefore,

    tan⁡−1[tan⁡(π4−θ)]=π4−θ\tan^{-1}\left[\tan\left(\frac{\pi}{4} - \theta\right)\right] = \frac{\pi}{4} - \theta

  7. Step 7: The equation becomes:

    π4−θ=θ2\frac{\pi}{4} - \theta = \frac{\theta}{2}

  8. Step 8: Solving:

    π4=3θ2\frac{\pi}{4} = \frac{3\theta}{2}

    which gives θ=π6\theta = \frac{\pi}{6}.

  9. Step 9: Since x=tan⁡θx = \tan\theta, we get:

    x=tan⁡(π6)=13x = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}

    ∴ x=13\boxed{x = \frac{1}{\sqrt{3}}}

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