If x + ² x = 1$\sin x + \sin^2 x = 1$, x in (0, (π)/(2))$x \in \left(0, \frac{\pi}{2}\right)$, then ( ¹² x + ¹² x) + 3 ( ¹⁰ x + ¹⁰ x + ⁸ x + ⁸ x) + ( ⁶ x + ⁶ x)$(\cos^{12} x + \tan^{12} x) + 3 (\cos^{10} x + \tan^{10} x + \cos^8 x + \tan^8 x) + (\cos^6 x + \tan^6 x)$ is equal to

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

Related Formula

Core Logic

Transform the trigonometric equation into a quadratic in terms of x$\sin x$:

Factorizing the quadratic:

Step 1: Finding bounds for a

Since x = 2$\sin x = 2$ has no real solution, we must have:

For this to have a solution, the root must lie in the standard domain of sine:

2 ≤ a ≤ 6$2 \le a \le 6$ Thus, the solution set is [p, q] = [2, 6]$[p, q] = [2, 6]$, meaning p = 2$p = 2$ and q = 6$q = 6$.

Step 2: Evaluating r

Evaluate r = 9° - 27° - 1 63° + 81°$r = \tan 9^{\circ} - \tan 27^{\circ} - \frac{1}{\cot 63^{\circ}} + \tan 81^{\circ}$. Using complementary angles ((90 - θ) = θ$\tan(90 - \theta) = \cot \theta$): 81° = 9°$\tan 81^{\circ} = \cot 9^{\circ}$ 1 63° = 63° = 27°$\frac{1}{\cot 63^{\circ}} = \tan 63^{\circ} = \cot 27^{\circ}$ Substitute these in:

Apply the formula θ + θ = (2)/( 2θ)$\tan \theta + \cot \theta = \frac{2}{\sin 2\theta}$:

We know 18° = √(5)-14$\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and 54° = 36° = √(5)+14$\sin 54^{\circ} = \cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$.

Step 3: Final Output Calculation

We need the value of pqr$pqr$:

Pattern Recognition

Converting mixed trig degrees like 9, 27, 63, 81 entirely into cot/tan pairs ALWAYS drops them into the (2)/( 2θ)$\frac{2}{\sin 2\theta}$ double-angle trap, bringing them natively to 18 and 54 degrees.

Chapter Mix

Class 11 Maths: Trigonometric Functions

Related Formula

If a θ + b θ = c$a\cos\theta + b\sin\theta = c$, dividing by θ$\cos\theta$ transforms the equation into a quadratic in terms of θ$\tan\theta$ and θ$\sec\theta$.

Core Logic

Given the equation:

Divide the entire equation by θ$\cos\theta$:

Square both sides to convert the θ$\sec\theta$ into a θ$\tan\theta$ expression:

Rearranging into a standard quadratic equation in terms of θ$\tan\theta$:

Step 1: Apply Quadratic Formula

Solve for θ$\tan\theta$ using the quadratic formula:

This gives two possible values:

Step 2: Check Extraneous Roots

When we squared the equation 4 + 5 θ = θ$4 + 5\tan\theta = \sec\theta$, we introduced the possibility of extraneous roots where θ$\sec\theta$ might be strictly negative while 4 + 5 θ$4 + 5\tan\theta$ is negative, but α in (-π/2, π/2)$\alpha \in (-\pi/2, \pi/2)$ restricts α gt 0$\cos\alpha \gt 0$, hence α gt 0$\sec\alpha \gt 0$. For α$\sec\alpha$ to be positive, 4 + 5 α gt 0$4 + 5\tan\alpha \gt 0$. If α = - 10 + √(10)12$\tan\alpha = -\frac{10 + \sqrt{10}}{12}$ (approx -1.09$-1.09$):

This contradicts α gt 0$\sec\alpha \gt 0$. Hence, this root is rejected.

Therefore, the only valid solution is:

Pattern Recognition

Whenever you square a trigonometric equation (like converting $\sec$ to $\tan$), always map the proposed roots back to the domain limits to prune out extraneous negative parity roots.

Chapter Mix

Class 11 Mathematics: Trigonometric Functions

Related Formula

Core Logic

Given equation:

Expanding both sides:

Step 1: Rearranging Terms

Grouping like terms together:

Dividing both sides by α β$\cos\alpha\cos\beta$:

Step 2: Conclusion

Comparing with the given equation α = k β$\tan\alpha = k\tan\beta$, we get k = -5$k = -5$.

Note by our answer (Bonus): Since α, β in (0, (π)/(2))$\alpha, \beta \in (0, \frac{\pi}{2})$, both α$\tan\alpha$ and β$\tan\beta$ must be positive. Hence, α = -5 β$\tan\alpha = -5\tan\beta$ is not possible. The data is inconsistent, but the NTA key marks option (2) as correct.

Pattern Recognition

Standard expansion of (A± B)$\sin(A\pm B)$ and grouping identical products to isolate (A)$\tan(A)$ and (B)$\tan(B)$.

Chapter Mix

Class 11 Maths: Trigonometric Functions