If k = tan left(fracpi4 +frac12cos^-1left(frac23right)right) + tan left(frac12sin^-1left(frac23right)right)$k = \tan \left(\frac{\pi}{4} +\frac{1}{2}\cos^{-1}\left(\frac{2}{3}\right)\right) + \tan \left(\frac{1}{2}\sin^{-1}\left(\frac{2}{3}\right)\right)$ then the number of solutions of the equation sin^-1(kx - 1) = sin^-1x - cos^-1x$\sin^{-1}(kx - 1) = \sin^{-1}x - \cos^{-1}x$ is ____.

### Related Formula sin^-1x + cos^-1x = fracpi2$$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$$ a^2 + b^2 = (a+b)^2 - 2ab$$a^2 + b^2 = (a+b)^2 - 2ab$$ ### Core Logic Let y = (sin^-1x)^2 + (cos^-1x)^2$y = (\sin^{-1}x)^2 + (\cos^{-1}x)^2$. y = (sin^-1x + cos^-1x)^2 - 2sin^-1x cos^-1x$$y = (\sin^{-1}x + \cos^{-1}x)^2 - 2\sin^{-1}x \cos^{-1}x$$ y = fracpi^24 - 2sin^-1x left(fracpi2 - sin^-1xright)$$y = \frac{\pi^2}{4} - 2\sin^{-1}x \left(\frac{\pi}{2} - \sin^{-1}x\right)$$ y = 2(sin^-1x)^2 - pi sin^-1x + fracpi^24$$y = 2(\sin^{-1}x)^2 - \pi \sin^{-1}x + \frac{\pi^2}{4}$$ Complete the square: y = 2left( sin^-1x - fracpi4 right)^2 - fracpi^28 + fracpi^24 = 2left( sin^-1x - fracpi4 right)^2 + fracpi^28$$y = 2\left( \sin^{-1}x - \frac{\pi}{4} \right)^2 - \frac{\pi^2}{8} + \frac{\pi^2}{4} = 2\left( \sin^{-1}x - \frac{\pi}{4} \right)^2 + \frac{\pi^2}{8}$$ ### Step 1: Bound the Function Given domain x in left[ -fracsqrt32, frac1sqrt2 right]$x \in \left[ -\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}} \right]$. The range of t = sin^-1x$t = \sin^{-1}x$ for this domain is left[ -fracpi3, fracpi4 right]$\left[ -\frac{\pi}{3}, \frac{\pi}{4} \right]$. So the expression is f(t) = 2left( t - fracpi4 right)^2 + fracpi^28$f(t) = 2\left( t - \frac{\pi}{4} \right)^2 + \frac{\pi^2}{8}$. ### Step 2: Find the Maximum The function f(t)$f(t)$ is a downward-opening distance squared logic? No, leading coefficient is positive, it's an upward parabola. Max value occurs at the boundary furthest from the vertex t = fracpi4$t = \frac{\pi}{4}$. The boundaries are -fracpi3$-\frac{\pi}{3}$ and fracpi4$\frac{\pi}{4}$. The distance from -fracpi3$-\frac{\pi}{3}$ to fracpi4$\frac{\pi}{4}$ is greatest. At t = -fracpi3$t = -\frac{\pi}{3}$: textMax = 2left( -fracpi3 - fracpi4 right)^2 + fracpi^28 = 2left( -frac7pi12 right)^2 + fracpi^28$$\text{Max} = 2\left( -\frac{\pi}{3} - \frac{\pi}{4} \right)^2 + \frac{\pi^2}{8} = 2\left( -\frac{7\pi}{12} \right)^2 + \frac{\pi^2}{8}$$ = 2left( frac49pi^2144 right) + fracpi^28 = frac49pi^272 + frac9pi^272 = frac58pi^272 = frac29pi^236$$= 2\left( \frac{49\pi^2}{144} \right) + \frac{\pi^2}{8} = \frac{49\pi^2}{72} + \frac{9\pi^2}{72} = \frac{58\pi^2}{72} = \frac{29\pi^2}{36}$$ ### Step 3: Calculate Required Value Here m = 29, n = 36$m = 29, n = 36$. Check gcd(29, 36) = 1$\gcd(29, 36) = 1$. m + n = 29 + 36 = 65$m + n = 29 + 36 = 65$. ### Pattern Recognition Expressions shaped like f(x)^2 + g(x)^2$f(x)^2 + g(x)^2$ where f(x)+g(x) = C$f(x)+g(x) = C$ will always map to a simple parabola. Analyze strictly based on vertex distance in the restricted f(x)$f(x)$ domain bounds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Inverse Trigonometric Functions Class 11 Maths: Quadratic Equations

### 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

### Related Formula tan(A - B) = fractan A - tan B1 + tan A tan B$$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$ tan 2theta = frac2tantheta1 - tan^2theta$$\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$$ ### Core Logic Let sin^-1frac2sqrt13 = theta$\sin^{-1}\frac{2}{\sqrt{13}} = \theta$ and cos^-1frac3sqrt10 = phi$\cos^{-1}\frac{3}{\sqrt{10}} = \phi$. Then sintheta = frac2sqrt13$\sin\theta = \frac{2}{\sqrt{13}}$ Rightarrow$\Rightarrow$ tantheta = frac23$\tan\theta = \frac{2}{3}$. And cosphi = frac3sqrt10$\cos\phi = \frac{3}{\sqrt{10}}$ Rightarrow$\Rightarrow$ tanphi = frac13$\tan\phi = \frac{1}{3}$. ### Execution Calculate tan 2theta$\tan 2\theta$: tan 2theta = frac2(2/3)1 - (4/9) = frac4/35/9 = frac125$$\tan 2\theta = \frac{2(2/3)}{1 - (4/9)} = \frac{4/3}{5/9} = \frac{12}{5}$$ Calculate tan 2phi$\tan 2\phi$: tan 2phi = frac2(1/3)1 - (1/9) = frac2/38/9 = frac68 = frac34$$\tan 2\phi = \frac{2(1/3)}{1 - (1/9)} = \frac{2/3}{8/9} = \frac{6}{8} = \frac{3}{4}$$ Now, substitute into the tan(2theta - 2phi)$\tan(2\theta - 2\phi)$ identity: tan(2theta - 2phi) = fracfrac125 - frac341 + left(frac125right)left(frac34right)$$\tan(2\theta - 2\phi) = \frac{\frac{12}{5} - \frac{3}{4}}{1 + \left(\frac{12}{5}\right)\left(\frac{3}{4}\right)}$$ = fracfrac48 - 15201 + frac3620 = fracfrac3320frac5620 = frac3356$$= \frac{\frac{48 - 15}{20}}{1 + \frac{36}{20}} = \frac{\frac{33}{20}}{\frac{56}{20}} = \frac{33}{56}$$ ### Pattern Recognition For compound inverse trigonometric expressions involving coefficients, map them entirely to their tan$\tan$ equivalents early to avoid cumbersome algebraic radicals. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Inverse Trigonometric Functions

### Related Formula cos(A + B) = cos A cos B - sin A sin B$$\cos(A + B) = \cos A \cos B - \sin A \sin B$$ sinleft(cos^-1 uright) = sqrt1 - u^2$$\sin\left(\cos^{-1} u\right) = \sqrt{1 - u^2}$$ ### Core Logic We expand the trigonometric compound angle expression to obtain a coupled algebraic equation relating variables x$x$ and y$y$. ### Step 1: Expand the equation using cosine addition formula Let theta = cos^-1left(fracx2right) implies cos theta = fracx2$\theta = \cos^{-1}\left(\frac{x}{2}\right) \implies \cos \theta = \frac{x}{2}$ and sin theta = sqrt1 - fracx^24$\sin \theta = \sqrt{1 - \frac{x^2}{4}}$: y = cosleft(fracpi3 + thetaright) = cosfracpi3 costheta - sinfracpi3 sintheta$$y = \cos\left(\frac{\pi}{3} + \theta\right) = \cos\frac{\pi}{3} \cos\theta - \sin\frac{\pi}{3} \sin\theta$$ y = frac12 left( fracx2 right) - fracsqrt32 sqrt1 - fracx^24$$y = \frac{1}{2} \left( \frac{x}{2} \right) - \frac{\sqrt{3}}{2} \sqrt{1 - \frac{x^2}{4}}$$ y = fracx4 - fracsqrt34 sqrt4 - x^2$$y = \frac{x}{4} - \frac{\sqrt{3}}{4} \sqrt{4 - x^2}$$ 4y = x - sqrt3sqrt4 - x^2$$4y = x - \sqrt{3}\sqrt{4 - x^2}$$ ### Step 2: Isolate the root and square Rearrange terms to isolate the radical and square both sides: x - 4y = sqrt3sqrt4 - x^2$$x - 4y = \sqrt{3}\sqrt{4 - x^2}$$ (x - 4y)^2 = 3(4 - x^2)$$(x - 4y)^2 = 3(4 - x^2)$$ x^2 - 8xy + 16y^2 = 12 - 3x^2$$x^2 - 8xy + 16y^2 = 12 - 3x^2$$ 4x^2 - 8xy + 16y^2 = 12$$4x^2 - 8xy + 16y^2 = 12$$ Divide the entire equation by 4: x^2 - 2xy + 4y^2 = 3$$x^2 - 2xy + 4y^2 = 3$$ ### Step 3: Evaluate the target expression We want to find the value of (x-y)^2 + 3y^2$(x-y)^2 + 3y^2$: (x-y)^2 + 3y^2 = x^2 - 2xy + y^2 + 3y^2 = x^2 - 2xy + 4y^2$$(x-y)^2 + 3y^2 = x^2 - 2xy + y^2 + 3y^2 = x^2 - 2xy + 4y^2$$ Notice that this matches the left side of our simplified equation from Step 2 exactly: (x - y)^2 + 3y^2 = 3$$(x - y)^2 + 3y^2 = 3$$ ### Pattern Recognition Coefficient symmetry: The expression (x-y)^2 + 3y^2 = x^2 - 2xy + 4y^2$(x-y)^2 + 3y^2 = x^2 - 2xy + 4y^2$ is a standard algebraic representation designed to match the quadratic expansion of scaled trigonometric sum equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Inverse Trigonometric Functions