( ⁻¹(3)/(5) + ⁻¹(5)/(13) + ⁻¹(33)/(65))$\cos \left(\sin^{-1}\frac{3}{5} +\sin^{-1}\frac{5}{13} +\sin^{-1}\frac{33}{65}\right)$ is equal to:

Related Formula

Core Logic

Let y = ( ⁻¹x)² + ( ⁻¹x)²$y = (\sin^{-1}x)^2 + (\cos^{-1}x)^2$.

Complete the square:

Step 1: Bound the Function

Given domain x in [ - √(3)2, 1√(2) ]$x \in \left[ -\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}} \right]$. The range of t = ⁻¹x$t = \sin^{-1}x$ for this domain is [ -(π)/(3), (π)/(4) ]$\left[ -\frac{\pi}{3}, \frac{\pi}{4} \right]$. So the expression is f(t) = 2( t - (π)/(4) )² + (π²)/(8)$f(t) = 2\left( t - \frac{\pi}{4} \right)^2 + \frac{\pi^2}{8}$.

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 = (π)/(4)$t = \frac{\pi}{4}$. The boundaries are -(π)/(3)$-\frac{\pi}{3}$ and (π)/(4)$\frac{\pi}{4}$. The distance from -(π)/(3)$-\frac{\pi}{3}$ to (π)/(4)$\frac{\pi}{4}$ is greatest. At t = -(π)/(3)$t = -\frac{\pi}{3}$:

Step 3: Calculate Required Value

Here m = 29, n = 36$m = 29, n = 36$. Check (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.

Chapter Mix

Class 12 Maths: Inverse Trigonometric Functions Class 11 Maths: Quadratic Equations

Related Formula

Core Logic

Given x in (- 12√(6), 12√(6))$x \in \left(-\frac{1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}}\right)$, the product (4x)(6x) = 24x² < 24((1)/(24)) = 1$(4x)(6x) = 24x^2 < 24\left(\frac{1}{24}\right) = 1$. Thus, the standard identity holds without additive phase shifts.

Step 1: Forming the Equation

Taking $\tan$ on both sides:

Cross multiply:

Step 2: Solving the Quadratic

Use the quadratic formula x = -b ± √(b² - 4ac)2a$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

Step 3: Validating Roots against Domain

We need to check which roots fall within (- 12√(6), 12√(6))$\left(-\frac{1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}}\right)$.

The positive root is x₁ = √(396) - 10√(3)48$x_1 = \frac{\sqrt{396} - 10\sqrt{3}}{48}$. Since √(396) > √(300) = 10√(3)$\sqrt{396} > \sqrt{300} = 10\sqrt{3}$, x₁ > 0$x_1 > 0$, making it a valid positive fraction.

The negative root is x₂ = -√(396) - 10√(3)48$x_2 = \frac{-\sqrt{396} - 10\sqrt{3}}{48}$. This is extremely negative and falls well outside the tight bound of - 12√(6)$-\frac{1}{2\sqrt{6}}$.

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.

Chapter Mix

Class 12 Maths: Inverse Trigonometric Functions Class 11 Maths: Quadratic Equations

Core Logic

First, evaluate k$k$. Let θ = (1)/(2) ⁻¹((2)/(3))$\theta = \frac{1}{2}\sin^{-1}\left(\frac{2}{3}\right)$. This implies (2θ) = (2)/(3)$\sin(2\theta) = \frac{2}{3}$. Since ⁻¹ x + ⁻¹ x = (π)/(2)$\cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}$, we have:

Substitute this back into k$k$:

Step 1: Simplify k

Multiply by 2/2$2/2$:

Since (2θ) = (2)/(3)$\sin(2\theta) = \frac{2}{3}$:

Step 2: Solve the Equation

Now solve ⁻¹(3x - 1) = ⁻¹x - ⁻¹x$\sin^{-1}(3x - 1) = \sin^{-1}x - \cos^{-1}x$. We know ⁻¹x = (π)/(2) - ⁻¹x$\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$.

Take sine of both sides:

Step 3: Finding Roots

Let ⁻¹x = α$\sin^{-1}x = \alpha$, so x = α$x = \sin\alpha$.

2x² - 3x = 0$2x^2 - 3x = 0$ x(2x - 3) = 0$x(2x - 3) = 0$ x = 0$x = 0$ or x = (3)/(2)$x = \frac{3}{2}$. Since domain of ⁻¹$\sin^{-1}$ is [-1, 1]$[-1, 1]$, x = 3/2$x = 3/2$ is rejected.

Now check x=0$x=0$ in original equation: LHS: ⁻¹(-1) = -π/2$\sin^{-1}(-1) = -\pi/2$ RHS: ⁻¹(0) - ⁻¹(0) = 0 - π/2 = -π/2$\sin^{-1}(0) - \cos^{-1}(0) = 0 - \pi/2 = -\pi/2$ Both sides match, so x=0$x=0$ is a valid solution. Wait, the official solution says x=0$x=0$ is rejected and number of solutions is 1? No, the snippet says "x=0, 3/2 (rejected)" which implies 3/2 is rejected. Then it says "No. of solution = 1". So x=0$x=0$ is indeed the 1 solution.

Chapter Mix

Class 12 Mathematics: Inverse Trigonometric Functions Class 11 Mathematics: Trigonometric Functions

Related Formula

Core Logic

Let ⁻¹ 2√(13) = θ$\sin^{-1}\frac{2}{\sqrt{13}} = \theta$ and ⁻¹ 3√(10) = φ$\cos^{-1}\frac{3}{\sqrt{10}} = \phi$. Then θ = 2√(13)$\sin\theta = \frac{2}{\sqrt{13}}$ ⇒$\Rightarrow$ θ = (2)/(3)$\tan\theta = \frac{2}{3}$. And φ = 3√(10)$\cos\phi = \frac{3}{\sqrt{10}}$ ⇒$\Rightarrow$ φ = (1)/(3)$\tan\phi = \frac{1}{3}$.

Execution

Calculate 2θ$\tan 2\theta$:

Calculate 2φ$\tan 2\phi$:

Now, substitute into the (2θ - 2φ)$\tan(2\theta - 2\phi)$ identity:

Pattern Recognition

For compound inverse trigonometric expressions involving coefficients, map them entirely to their $\tan$ equivalents early to avoid cumbersome algebraic radicals.

Chapter Mix

Class 12 Maths: Inverse Trigonometric Functions

Related Formula

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 θ = ⁻¹((x)/(2)) θ = (x)/(2)$\theta = \cos^{-1}\left(\frac{x}{2}\right) \implies \cos \theta = \frac{x}{2}$ and θ = √(1 - (x²)/(4))$\sin \theta = \sqrt{1 - \frac{x^2}{4}}$:

Step 2: Isolate the root and square

Rearrange terms to isolate the radical and square both sides:

Divide the entire equation by 4:

Step 3: Evaluate the target expression

We want to find the value of (x-y)² + 3y²$(x-y)^2 + 3y^2$:

Notice that this matches the left side of our simplified equation from Step 2 exactly:

Pattern Recognition

Coefficient symmetry: The expression (x-y)² + 3y² = x² - 2xy + 4y²$(x-y)^2 + 3y^2 = x^2 - 2xy + 4y^2$ is a standard algebraic representation designed to match the quadratic expansion of scaled trigonometric sum equations.

Chapter Mix

Class 12 Mathematics: Inverse Trigonometric Functions