Question:

Solve for x, given \( \tan^{-1}\left(\frac{1 - x}{1 + x}\right) = \frac{1}{2} \tan^{-1}(x) \)

Updated On: Apr 13, 2025
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Solution and Explanation

Given Equation:
We are given the equation: \[ \tan^{-1}\left(\frac{1 - x}{1 + x}\right) = \frac{1}{2} \tan^{-1}(x) \]

Step 1: Use the Double Angle Identity for Tangent:
Recall that the identity for the tangent of a half-angle is: \[ \tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{1 + \cos(\theta)} \] To apply this identity, let \( \theta = \tan^{-1}(x) \), so that: \[ \frac{1 - x}{1 + x} = \tan\left(\frac{1}{2} \tan^{-1}(x)\right) \] Now, the left-hand side is \( \tan^{-1} \left( \frac{1 - x}{1 + x} \right) \), and we can equate the two sides: \[ \tan^{-1} \left( \frac{1 - x}{1 + x} \right) = \frac{1}{2} \tan^{-1}(x) \]

Step 2: Make a Substitution:
Let \( y = \tan^{-1}(x) \), so that \( x = \tan(y) \). Now the equation becomes: \[ \tan^{-1} \left( \frac{1 - \tan(y)}{1 + \tan(y)} \right) = \frac{y}{2} \]

Step 3: Simplify and Solve for \( x \):
Recognize that the left-hand side simplifies using the identity for tangent subtraction: \[ \frac{1 - \tan(y)}{1 + \tan(y)} = \tan\left(\frac{\pi}{4} - y\right) \] Thus, the equation becomes: \[ \tan\left(\frac{\pi}{4} - y\right) = \frac{y}{2} \] Solving this equation by trial or approximation, we find that: \[ x = \sqrt{3} \]

Final Answer:
Therefore, the solution for \( x \) is: \[ x = \sqrt{3} \]

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Concepts Used:

Properties of Inverse Trigonometric Functions

The elementary properties of inverse trigonometric functions will help to solve problems. Here are a few important properties related to inverse trigonometric functions:

Property Set 1:

  • Sin−1(x) = cosec−1(1/x), x∈ [−1,1]−{0}
  • Cos−1(x) = sec−1(1/x), x ∈ [−1,1]−{0}
  • Tan−1(x) = cot−1(1/x), if x > 0  (or)  cot−1(1/x) −π, if x < 0
  • Cot−1(x) = tan−1(1/x), if x > 0 (or) tan−1(1/x) + π, if x < 0

Property Set 2:

  • Sin−1(−x) = −Sin−1(x)
  • Tan−1(−x) = −Tan−1(x)
  • Cos−1(−x) = π − Cos−1(x)
  • Cosec−1(−x) = − Cosec−1(x)
  • Sec−1(−x) = π − Sec−1(x)
  • Cot−1(−x) = π − Cot−1(x)

Property Set 3:

  • Sin−1(1/x) = cosec−1x, x≥1 or x≤−1
  • Cos−1(1/x) = sec−1x, x≥1 or x≤−1
  • Tan−1(1/x) = −π + cot−1(x)

Property Set 4:

  • Sin−1(cos θ) = π/2 − θ, if θ∈[0,π]
  • Cos−1(sin θ) = π/2 − θ, if θ∈[−π/2, π/2]
  • Tan−1(cot θ) = π/2 − θ, θ∈[0,π]
  • Cot−1(tan θ) = π/2 − θ, θ∈[−π/2, π/2]
  • Sec−1(cosec θ) = π/2 − θ, θ∈[−π/2, 0]∪[0, π/2]
  • Cosec−1(sec θ) = π/2 − θ, θ∈[0,π]−{π/2}
  • Sin−1(x) = cos−1[√(1−x2)], 0≤x≤1 = −cos−1[√(1−x2)], −1≤x<0

Property Set 5:

  • Sin−1x + Cos−1x = π/2
  • Tan−1x + Cot−1(x) = π/2
  • Sec−1x + Cosec−1x = π/2

Property Set 6:

  • If x, y > 0

Tan−1x + Tan−1y = π + tan−1 (x+y/ 1-xy), if xy > 1

Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1

  • If x, y < 0

Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1

Tan−1x + Tan−1y = -π + tan−1 (x+y/ 1-xy), if xy > 1

Property Set 7:

  • sin−1(x) + sin−1(y) = sin−1[x√(1−y2)+ y√(1−x2)]
  • cos−1x + cos−1y = cos−1[xy−√(1−x2)√(1−y2)]

Property Set 8:

  • sin−1(sin x) = −π−π, if x∈[−3π/2, −π/2]

= x, if x∈[−π/2, π/2]

= π−x, if x∈[π/2, 3π/2]

=−2π+x, if x∈[3π/2, 5π/2] And so on.

  • cos−1(cos x) = 2π+x, if x∈[−2π,−π]

= −x, ∈[−π,0]

= x, ∈[0,π]

= 2π−x, ∈[π,2π]

=−2π+x, ∈[2π,3π]

  • tan−1(tan x) = π+x, x∈(−3π/2, −π/2)

= x, (−π/2, π/2)

= x−π, (π/2, 3π/2)

= x−2π, (3π/2, 5π/2)

Property Set 9: