If tan⁻¹ 2x + tan⁻¹ 3x = π/4, then: Prove that 6x² + 5x - 1 = 0. Hence, solve the equation tan⁻¹ 2x…

If $\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4}$, then:

(a)[2]

Prove that $6x^2 + 5x - 1 = 0$.

(b)[2]

Hence, solve the equation $\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4}$.

Answer

Answer (a)

From $\tan^{-1}2x+\tan^{-1}3x=\frac{\pi}{4}$ we get $6x^2+5x-1=0$.

1. Given $\tan^{-1}2x+\tan^{-1}3x=\frac{\pi}{4}$.
2. Using $\tan^{-1}a+\tan^{-1}b=\tan^{-1}\frac{a+b}{1-ab}$ (for $ab<1$): $\tan^{-1}\frac{2x+3x}{1-6x^2}=\frac{\pi}{4}$.
3. Take tangent of both sides: $\frac{5x}{1-6x^2}=\tan\frac{\pi}{4}=1$.
4. Cross-multiply: $5x=1-6x^2$.
5. Rearrange: $6x^2+5x-1=0$.

Answer (b)

Factorise $6x^2+5x-1=(6x-1)(x+1)=0$, so $x=\frac{1}{6}$ or $x=-1$. For $x=-1$: $\tan^{-1}(-2)+\tan^{-1}(-3)$ is negative (about $-\frac{3\pi}{4}$), not $\frac{\pi}{4}$ (here $6x^2=6>1$, so the formula does not apply). Rejected. For $x=\frac{1}{6}$: $\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{1}{2}=\tan^{-1}\frac{5/6}{5/6}=\frac{\pi}{4}$. Accepted.

Final answer: $\frac{1}{6}$