The number of solutions of tan^-14x + tan^-16x = fracpi6$\tan^{-1}4x + \tan^{-1}6x = \frac{\pi}{6}$, where -frac12sqrt6 < x < frac12sqrt6$-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$ is equal to

### Related Formula tan^-1a + tan^-1b = tan^-1left(fraca + b1 - abright) quad textfor ab < 1$$\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \quad \text{for } ab < 1$$ ### Core Logic Given x in left(-frac12sqrt6, frac12sqrt6right)$x \in \left(-\frac{1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}}\right)$, the product (4x)(6x) = 24x^2 < 24left(frac124right) = 1$(4x)(6x) = 24x^2 < 24\left(\frac{1}{24}\right) = 1$. Thus, the standard identity holds without additive phase shifts. tan^-14x + tan^-16x = fracpi6$$\tan^{-1}4x + \tan^{-1}6x = \frac{\pi}{6}$$ tan^-1left(frac4x + 6x1 - 24x^2right) = fracpi6$$\tan^{-1}\left(\frac{4x + 6x}{1 - 24x^2}\right) = \frac{\pi}{6}$$ ### Step 1: Forming the Equation Taking tan$\tan$ on both sides: frac10x1 - 24x^2 = tanleft(fracpi6right) = frac1sqrt3$$\frac{10x}{1 - 24x^2} = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$$ Cross multiply: 10sqrt3 x = 1 - 24x^2$$10\sqrt{3} x = 1 - 24x^2$$ 24x^2 + 10sqrt3 x - 1 = 0$$24x^2 + 10\sqrt{3} x - 1 = 0$$

### Step 2: Solving the Quadratic Use the quadratic formula x = frac-b pm sqrtb^2 - 4ac2a$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: x = frac-10sqrt3 pm sqrt(10sqrt3)^2 - 4(24)(-1)2(24)$$x = \frac{-10\sqrt{3} \pm \sqrt{(10\sqrt{3})^2 - 4(24)(-1)}}{2(24)}$$ x = frac-10sqrt3 pm sqrt300 + 9648$$x = \frac{-10\sqrt{3} \pm \sqrt{300 + 96}}{48}$$ x = frac-10sqrt3 pm sqrt39648$$x = \frac{-10\sqrt{3} \pm \sqrt{396}}{48}$$ ### Step 3: Validating Roots against Domain We need to check which roots fall within left(-frac12sqrt6, frac12sqrt6right)$\left(-\frac{1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}}\right)$. The positive root is x_1 = fracsqrt396 - 10sqrt348$x_1 = \frac{\sqrt{396} - 10\sqrt{3}}{48}$. Since sqrt396 > sqrt300 = 10sqrt3$\sqrt{396} > \sqrt{300} = 10\sqrt{3}$, x_1 > 0$x_1 > 0$, making it a valid positive fraction. The negative root is x_2 = frac-sqrt396 - 10sqrt348$x_2 = \frac{-\sqrt{396} - 10\sqrt{3}}{48}$. This is extremely negative and falls well outside the tight bound of -frac12sqrt6$-\frac{1}{2\sqrt{6}}$. Only 1 solution exists in the specified domain. ### Pattern Recognition Checking the valid domain for ab < 1$ab < 1$ prevents phantom solutions. A quadratic always yields two roots, but the strict interval provided in the question stems directly from the convergence limits of the inverse tangent addition identity, aggressively discarding the far-flung negative root. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Inverse Trigonometric Functions Class 11 Maths: Quadratic Equations