Solve for x: $\tan^{- 1} 3 x + \tan^{- 1} 2 x = \frac{π}{4}$

Solution

$\tan^{- 1} 3 x + \tan^{- 1} 2 x = \frac{π}{4} \Rightarrow \tan^{- 1} (\frac{5 x}{1 - 6 x^{2}}) = \frac{π}{4}, 3 x \times 2 x < 1 \Rightarrow \tan [\tan^{- 1} (\frac{5 x}{1 - 6 x^{2}})] = \tan \frac{π}{4} \Rightarrow \frac{5 x}{1 - 6 x^{2}} = 1 \Rightarrow 1 - 6 x^{2} = 5 x \Rightarrow 6 x^{2} + 5 x - 1 = 0 \Rightarrow 6 x^{2} + 6 x - x - 1 = 0 \Rightarrow 6 x (x + 1) - 1 (x + 1) = 0 \Rightarrow (6 x - 1) (x + 1) = 0 \Rightarrow 6 x - 1 = 0 or x + 1 = 0 \Rightarrow x = \frac{1}{6} or x = - 1 Here (- 3) x (- 2) ≮ 1 [∵ (- 3) x (- 2) = 6 > 1]$

Therefore, x= -1 is not the solution.

When substituting x = $\frac{1}{6}$ in 3x $\times$ 2x, we have,

$3 x \frac{1}{6} x 2 x \frac{1}{6} = \frac{1}{2} x \frac{1}{3} = \frac{1}{6} < 1 .$

Hence x = $\frac{1}{6}$ is the solution of the given equation.

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Inverse Trigonometric Functions

Solve for x: $\tan^{- 1} 3 x + \tan^{- 1} 2 x = \frac{π}{4}$

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Inverse Trigonometric Functions

Solve for x: tan-1 3x + tan-1 2x = π4

Some More Questions From Inverse Trigonometric Functions Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Solve for x: $\tan^{- 1} 3 x + \tan^{- 1} 2 x = \frac{π}{4}$