If 10 ⁴θ + 15 ⁴θ = 6$10\sin^4\theta + 15\cos^4\theta = 6$, then the value of (27 ⁶θ + 8 ⁶θ)/(16 ⁸θ)$\frac{27\csc^6\theta + 8\sec^6\theta}{16\sec^8\theta}$ is:

Related Formula

Apply the fundamental trigonometric identity properties directly linking reciprocal functions:

Core Logic

Simplify the given trigonometric equation using the identity formulas:

Substitute these simplified expressions back into the target relation equation:

Step 1: Set up a Quadratic Equation for the roots

We are given the linear \sum α + β = 8$\alpha + \beta = 8$. Use the algebraic identity for squares to find the product:

Since we know both the \sum (8) and product (15), α$\alpha$ and \β$\beta$ are the roots of the quadratic equation:

Step 2: Assign Variables and Compute the Target Value

Using the given constraint condition α ≤ β$\alpha \leq \beta$, we assign the values as:

Now substitute these values into the evaluation expression:

Pattern Recognition

Recognizing standard algebraic forms for sums and products like α+β$\alpha+\beta$ and αβ$\alpha\beta$ helps identify the system's values without needing to use full square root quadratic formulas.

Chapter Mix

Class 12 Mathematics: Inverse Trigonometric Functions

Related Formula

Identity for reciprocal of product of sines with an arithmetic progression phase difference β$\beta$:

where β = A - B$\beta = A - B$.

Core Logic

Let θᵣ = (π)/(4) + r(π)/(6)$\theta_r = \frac{\pi}{4} + r\frac{\pi}{6}$. Then the difference between consecutive angles is:

Multiply and divide the general term of the summation by ((π)/(6))$\sin\left(\frac{\pi}{6}\right)$:

Step 1: Expand the Telescopic Sum

Where:

Now compute ((23π)/(12)) = (2π - (π)/(12)) = - ((π)/(12))$\cot\left(\frac{23\pi}{12}\right) = \cot\left(2\pi - \frac{\pi}{12}\right) = -\cot\left(\frac{\pi}{12}\right)$:

Step 2: Solve for a and b

Accounting for the absolute value ranges across the quadrants and simplifying the telescopic intervals:

Comparing with a√(3) + b$a\sqrt{3} + b$:

Now calculate a² + b²$a^2 + b^2$:

Pattern Recognition

Telescopic series involving absolute values of trigonometric products require careful tracking of interval signs across quadrants before applying boundary difference reductions.

Chapter Mix

Class 11 Mathematics: Trigonometric Functions

Related Formula

Core Logic

Rearrange using the principal trigonometric identity property ⁻¹ x = (π)/(2) - ⁻¹ x$\sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x$:

Step 1: Isolate angles using variable substitution

Let ⁻¹ x = α$\cos^{-1} x = \alpha$ and ⁻¹(2x + 1) = β$\sin^{-1}(2x + 1) = \beta$.

Take cosine on both sides:

Step 2: Convert to algebraic identity form

Using double \angle formulas:

Since α = x$\cos\alpha = x$ and β = 2x + 1$\sin\beta = 2x + 1$, substitute directly:

Solving the quadratic equation gives:

Checking domain validation constraints for inverse functions dictates that only x = 1 - √(5)2$x = \frac{1 - \sqrt{5}}{2}$ is valid (the positive root exceeds principal bounds).

Step 3: Evaluate target question calculation

From the valid root, we track:

Squaring both sides yields:

Pattern Recognition

When inverse sums equate to values like (3π)/(2)$\frac{3\pi}{2}$, look for extreme boundary values or domain constraints to eliminate invalid algebraic roots.

Chapter Mix

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

Related Formula

Trigonometric Addition Identity for tangent:

Core Logic

Given expressions for A$\tan A$ and B$\tan B$:

Step 1: Simplify the Compound Tangent Formula

Simplify the numerator:

Simplify the denominator:

Now put them together:

Step 2: Compare with tan C

Cancelling out (1+x)$(1+x)$ and matching root expressions:

Now evaluate C$\tan C$:

Since (A+B) = C$\tan(A+B) = \tan C$ and both arguments are in acute range:

A+B = C$A+B = C$

Pattern Recognition

Sees: Multi-variable algebraic rational terms involving square roots. Shortcut: If algebraic tracking feels complicated, substitute a simple valid number like x=1$x=1$ to evaluate coefficients dynamically: A = 1√(3)$\tan A = \frac{1}{\sqrt{3}}$, B = 1√(3) A=30°, B=30° A+B=60°$\tan B = \frac{1}{\sqrt{3}} \implies A=30^{\circ}, B=30^{\circ} \implies A+B=60^{\circ}$. Then C = √(1+1+1) = √(3) C=60°$\tan C = \sqrt{1+1+1} = \sqrt{3} \implies C=60^{\circ}$. Thus A+B=C$A+B=C$ holds instantly.

Chapter Mix

Class 11 Mathematics: Trigonometric Functions Class 10 Mathematics: Algebraic Identities

Related Formula

Core Logic

Let us simplify the LHS expression:

Assuming 2x ≠ 0$\cos 2x \neq 0$:

Since ² 2x = 1 - ² 2x$\sin^2 2x = 1 - \cos^2 2x$, the denominator becomes:

Therefore:

Step 1: Solving the Algebraic Equation

Equating LHS to RHS:

By inspection, x = -1$x = -1$ is a root:

Factoring out (x + 1)$(x + 1)$:

For the quadratic factor x² - 2x + 2 = 0$x^2 - 2x + 2 = 0$, the discriminant is D = (-2)² - 4(1)(2) = -4 < 0$D = (-2)^2 - 4(1)(2) = -4 < 0$, yielding no real roots. Thus, the only real solution is x = -1$x = -1$, and its sum is -1$-1$.

Pattern Recognition

Complicated mixed expressions of trigonometric fractions often collapse into simple constants upon identity transformations. Look for factorization templates like a³ - b³$a^3 - b^3$ or a⁶ - b⁶$a^6 - b^6$.

Chapter Mix

Class 11 Mathematics: Trigonometric Functions Class 12 Mathematics: Polynomial Equations