Solve for real x: tan^-1x+tan^-1(2x)=π/4. State all solutions.

Solve for real $x$ : $tan^{- 1} x + tan^{- 1} (2 x) = \frac{π}{4}$ . State all solutions.

hard 3 marks Class 12 · Mathematics · Inverse Trigonometric Functions

Answer

x = \frac{- 3 p m s q r t 17}{4}

The correct answer is $x = \frac{- 3 p m s q r t 17}{4}$ — Both values are admissible because $tan^{- 1}$ is defined for all real $x$ .

Let $A = tan^{- 1} x$ and $B = tan^{- 1} (2 x)$ . Then $tan A = x$ and $tan B = 2 x$ .
Using $tan (A + B) = \frac{t a n A + t a n B}{1 - t a n A t a n B}$ , we get $tan! (\frac{π}{4}) = 1 = \frac{x + 2 x}{, 1 - 2 x ^{2} ,} = \frac{3 x}{1 - 2 x ^{2}}$ . So $\frac{3 x}{1 - 2 x ^{2}} = 1$ .
Cross‑multiplying: $3 x = 1 - 2 x^{2}$ . Rearranging gives $2 x^{2} + 3 x - 1 = 0$ . Solve the quadratic: $x = \frac{- 3 p m s q r t 9 + 8}{4} = \frac{- 3 p m s q r t 17}{4}$ .
Both values are admissible because $tan^{- 1}$ is defined for all real $x$ .

Common mistake: Ignoring the domain of arctan or forgetting the plus/minus in quadratic

Concept tested: Equation with sum of arctangents