### Related Formula Trigonometric Addition Identity for tangent: tan(A+B) = fractan A + tan B1 - tan A tan B$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ ### Core Logic Given expressions for tan A$\tan A$ and tan B$\tan B$: tan(A+B) = fracfrac1sqrtx(x^2+x+1) + fracsqrtxsqrtx^2+x+11 - left(frac1sqrtx(x^2+x+1)right) cdot left(fracsqrtxsqrtx^2+x+1right) $\tan(A+B) = \frac{\frac{1}{\sqrt{x(x^2+x+1)}} + \frac{\sqrt{x}}{\sqrt{x^2+x+1}}}{1 - \left(\frac{1}{\sqrt{x(x^2+x+1)}}\right) \cdot \left(\frac{\sqrt{x}}{\sqrt{x^2+x+1}}\right)} $ ### Step 1: Simplify the Compound Tangent Formula Simplify the numerator: textNumerator = frac1 + xsqrtxsqrtx^2+x+1$\text{Numerator} = \frac{1 + x}{\sqrt{x}\sqrt{x^2+x+1}}$ Simplify the denominator: textDenominator = 1 - frac1x^2+x+1 = fracx^2+x+1-1x^2+x+1 = fracx^2+xx^2+x+1 = fracx(x+1)x^2+x+1$\text{Denominator} = 1 - \frac{1}{x^2+x+1} = \frac{x^2+x+1-1}{x^2+x+1} = \frac{x^2+x}{x^2+x+1} = \frac{x(x+1)}{x^2+x+1}$ Now put them together: tan(A+B) = fracfrac1+xsqrtxsqrtx^2+x+1fracx(x+1)x^2+x+1 = frac(1+x)(x^2+x+1)sqrtxsqrtx^2+x+1 cdot x(x+1) $\tan(A+B) = \frac{\frac{1+x}{\sqrt{x}\sqrt{x^2+x+1}}}{\frac{x(x+1)}{x^2+x+1}} = \frac{(1+x)(x^2+x+1)}{\sqrt{x}\sqrt{x^2+x+1} \cdot x(x+1)} $ ### Step 2: Compare with tan C Cancelling out (1+x)$(1+x)$ and matching root expressions: tan(A+B) = fracsqrtx^2+x+1xsqrtx $\tan(A+B) = \frac{\sqrt{x^2+x+1}}{x\sqrt{x}} $ Now evaluate tan C$\tan C$: tan C = sqrtfrac1x^3 + frac1x^2 + frac1x = sqrtfrac1+x+x^2x^3 = fracsqrtx^2+x+1xsqrtx $\tan C = \sqrt{\frac{1}{x^3} + \frac{1}{x^2} + \frac{1}{x}} = \sqrt{\frac{1+x+x^2}{x^3}} = \frac{\sqrt{x^2+x+1}}{x\sqrt{x}} $ Since tan(A+B) = tan 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: tan A = frac1sqrt3$\tan A = \frac{1}{\sqrt{3}}$, tan B = frac1sqrt3 implies A=30^circ, B=30^circ implies A+B=60^circ$\tan B = \frac{1}{\sqrt{3}} \implies A=30^{\circ}, B=30^{\circ} \implies A+B=60^{\circ}$. Then tan C = sqrt1+1+1 = sqrt3 implies C=60^circ$\tan C = \sqrt{1+1+1} = \sqrt{3} \implies C=60^{\circ}$. Thus A+B=C$A+B=C$ holds instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometric Functions Class 10 Mathematics: Algebraic Identities

### Related Formula cos^6 x - sin^6 x = (cos^2 x - sin^2 x)(cos^4 x + cos^2 x sin^2 x + sin^4 x) = cos 2x (1 - sin^2 x cos^2 x)$\cos^6 x - \sin^6 x = (\cos^2 x - \sin^2 x)(\cos^4 x + \cos^2 x \sin^2 x + \sin^4 x) = \cos 2x (1 - \sin^2 x \cos^2 x)$ ### Core Logic Let us simplify the LHS expression: textLHS = fraccos 2x (3 + cos^2 2x)cos 2x (1 - sin^2 x cos^2 x)$\text{LHS} = \frac{\cos 2x (3 + \cos^2 2x)}{\cos 2x (1 - \sin^2 x \cos^2 x)}$ Assuming cos 2x neq 0$\cos 2x \neq 0$: textLHS = frac3 + cos^2 2x1 - frac14sin^2 2x = frac4(3 + cos^2 2x)4 - sin^2 2x$\text{LHS} = \frac{3 + \cos^2 2x}{1 - \frac{1}{4}\sin^2 2x} = \frac{4(3 + \cos^2 2x)}{4 - \sin^2 2x}$ Since sin^2 2x = 1 - cos^2 2x$\sin^2 2x = 1 - \cos^2 2x$, the denominator becomes: 4 - (1 - cos^2 2x) = 3 + cos^2 2x$4 - (1 - \cos^2 2x) = 3 + \cos^2 2x$ Therefore: textLHS = frac4(3 + cos^2 2x)3 + cos^2 2x = 4$\text{LHS} = \frac{4(3 + \cos^2 2x)}{3 + \cos^2 2x} = 4$ ### Step 1: Solving the Algebraic Equation Equating LHS to RHS: 4 = x^3 - x^2 + 6 implies x^3 - x^2 + 2 = 0$4 = x^3 - x^2 + 6 \implies x^3 - x^2 + 2 = 0$ By inspection, x = -1$x = -1$ is a root: (-1)^3 - (-1)^2 + 2 = -1 - 1 + 2 = 0$(-1)^3 - (-1)^2 + 2 = -1 - 1 + 2 = 0$ Factoring out (x + 1)$(x + 1)$: (x + 1)(x^2 - 2x + 2) = 0$(x + 1)(x^2 - 2x + 2) = 0$ For the quadratic factor x^2 - 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 a^3 - b^3$a^3 - b^3$ or a^6 - b^6$a^6 - b^6$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometric Functions Class 12 Mathematics: Polynomial Equations

### Related Formula cos 2x = 1 - 2sin^2 x$\cos 2x = 1 - 2\sin^2 x$ tan theta + cot theta = frac2sin 2theta$\tan \theta + \cot \theta = \frac{2}{\sin 2\theta}$ ### Core Logic Transform the trigonometric equation into a quadratic in terms of sin x$\sin x$: (1 - 2sin^2 x) + asin x = 2a - 7$(1 - 2\sin^2 x) + a\sin x = 2a - 7$ 2sin^2 x - asin x + 2a - 8 = 0$2\sin^2 x - a\sin x + 2a - 8 = 0$ Factorizing the quadratic: 2sin^2 x - 4sin x - (a-4)sin x + 2(a-4) = 0$2\sin^2 x - 4\sin x - (a-4)\sin x + 2(a-4) = 0$ 2sin x(sin x - 2) - (a-4)(sin x - 2) = 0$2\sin x(\sin x - 2) - (a-4)(\sin x - 2) = 0$ (sin x - 2)(2sin x - (a-4)) = 0$(\sin x - 2)(2\sin x - (a-4)) = 0$ ### Step 1: Finding bounds for a Since sin x = 2$\sin x = 2$ has no real solution, we must have: sin x = fraca-42$\sin x = \frac{a-4}{2}$ For this to have a solution, the root must lie in the standard domain of sine: -1 le fraca-42 le 1$-1 \le \frac{a-4}{2} \le 1$ -2 le a-4 le 2$-2 \le a-4 \le 2$ 2 le a le 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 = tan 9^circ - tan 27^circ - frac1cot 63^circ + tan 81^circ$r = \tan 9^{\circ} - \tan 27^{\circ} - \frac{1}{\cot 63^{\circ}} + \tan 81^{\circ}$. Using complementary angles (tan(90 - theta) = cot theta$\tan(90 - \theta) = \cot \theta$): tan 81^circ = cot 9^circ$\tan 81^{\circ} = \cot 9^{\circ}$ frac1cot 63^circ = tan 63^circ = cot 27^circ$\frac{1}{\cot 63^{\circ}} = \tan 63^{\circ} = \cot 27^{\circ}$ Substitute these in: r = (tan 9^circ + cot 9^circ) - (tan 27^circ + cot 27^circ)$r = (\tan 9^{\circ} + \cot 9^{\circ}) - (\tan 27^{\circ} + \cot 27^{\circ})$ Apply the formula tan theta + cot theta = frac2sin 2theta$\tan \theta + \cot \theta = \frac{2}{\sin 2\theta}$: r = frac2sin 18^circ - frac2sin 54^circ$r = \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$ We know sin 18^circ = fracsqrt5-14$\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and sin 54^circ = cos 36^circ = fracsqrt5+14$\sin 54^{\circ} = \cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$. r = frac8sqrt5-1 - frac8sqrt5+1 = 8 left[ fracsqrt5+1 - (sqrt5-1)(sqrt5-1)(sqrt5+1) right]$r = \frac{8}{\sqrt{5}-1} - \frac{8}{\sqrt{5}+1} = 8 \left[ \frac{\sqrt{5}+1 - (\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)} \right]$ r = 8 left[ frac24 right] = 4$r = 8 \left[ \frac{2}{4} \right] = 4$ ### Step 3: Final Output Calculation We need the value of pqr$pqr$: pqr = 2 times 6 times 4 = 48$pqr = 2 \times 6 \times 4 = 48$ ### Pattern Recognition Converting mixed trig degrees like 9, 27, 63, 81 entirely into cot/tan pairs ALWAYS drops them into the frac2sin 2theta$\frac{2}{\sin 2\theta}$ double-angle trap, bringing them natively to 18 and 54 degrees. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Trigonometric Functions

### Related Formula sec^2theta - tan^2theta = 1$\sec^2\theta - \tan^2\theta = 1$ If acostheta + bsintheta = c$a\cos\theta + b\sin\theta = c$, dividing by costheta$\cos\theta$ transforms the equation into a quadratic in terms of tantheta$\tan\theta$ and sectheta$\sec\theta$. ### Core Logic Given the equation: 4costheta + 5sintheta = 1$4\cos\theta + 5\sin\theta = 1$ Divide the entire equation by costheta$\cos\theta$: 4 + 5tantheta = sectheta$4 + 5\tan\theta = \sec\theta$ Square both sides to convert the sectheta$\sec\theta$ into a tantheta$\tan\theta$ expression: (4 + 5tantheta)^2 = sec^2theta$(4 + 5\tan\theta)^2 = \sec^2\theta$ 16 + 25tan^2theta + 40tantheta = 1 + tan^2theta$16 + 25\tan^2\theta + 40\tan\theta = 1 + \tan^2\theta$ Rearranging into a standard quadratic equation in terms of tantheta$\tan\theta$: 24tan^2theta + 40tantheta + 15 = 0$24\tan^2\theta + 40\tan\theta + 15 = 0$ ### Step 1: Apply Quadratic Formula Solve for tantheta$\tan\theta$ using the quadratic formula: tantheta = frac-40 pm sqrt1600 - 4(24)(15)2(24)$\tan\theta = \frac{-40 \pm \sqrt{1600 - 4(24)(15)}}{2(24)}$ tantheta = frac-40 pm sqrt1600 - 144048$\tan\theta = \frac{-40 \pm \sqrt{1600 - 1440}}{48}$ tantheta = frac-40 pm sqrt16048$\tan\theta = \frac{-40 \pm \sqrt{160}}{48}$ tantheta = frac-40 pm 4sqrt1048$\tan\theta = \frac{-40 \pm 4\sqrt{10}}{48}$ tantheta = frac-10 pm sqrt1012$\tan\theta = \frac{-10 \pm \sqrt{10}}{12}$ This gives two possible values: tantheta = frac-10 + sqrt1012 quad textand quad tantheta = -left(frac10 + sqrt1012right)$\tan\theta = \frac{-10 + \sqrt{10}}{12} \quad \text{and} \quad \tan\theta = -\left(\frac{10 + \sqrt{10}}{12}\right)$ ### Step 2: Check Extraneous Roots When we squared the equation 4 + 5tantheta = sectheta$4 + 5\tan\theta = \sec\theta$, we introduced the possibility of extraneous roots where sectheta$\sec\theta$ might be strictly negative while 4 + 5tantheta$4 + 5\tan\theta$ is negative, but alpha in (-pi/2, pi/2)$\alpha \in (-\pi/2, \pi/2)$ restricts cosalpha gt 0$\cos\alpha \gt 0$, hence secalpha gt 0$\sec\alpha \gt 0$. For secalpha$\sec\alpha$ to be positive, 4 + 5tanalpha gt 0$4 + 5\tan\alpha \gt 0$. If tanalpha = -frac10 + sqrt1012$\tan\alpha = -\frac{10 + \sqrt{10}}{12}$ (approx -1.09$-1.09$): 4 + 5(-1.09) = 4 - 5.45 = -1.45 lt 0$4 + 5(-1.09) = 4 - 5.45 = -1.45 \lt 0$ This contradicts secalpha gt 0$\sec\alpha \gt 0$. Hence, this root is rejected. Therefore, the only valid solution is: tanalpha = fracsqrt10 - 1012$\tan\alpha = \frac{\sqrt{10} - 10}{12}$ ### Pattern Recognition Whenever you square a trigonometric equation (like converting sec$\sec$ to tan$\tan$), always map the proposed roots back to the domain limits to prune out extraneous negative parity roots. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometric Functions

### Related Formula sin(A pm B) = sin A cos B pm cos A sin B$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$ ### Core Logic Given equation: 3sin(alpha + beta) = 2sin(alpha - beta)$3\sin(\alpha + \beta) = 2\sin(\alpha - \beta)$ Expanding both sides: 3(sinalphacosbeta + cosalphasinbeta) = 2(sinalphacosbeta - cosalphasinbeta)$3(\sin\alpha\cos\beta + \cos\alpha\sin\beta) = 2(\sin\alpha\cos\beta - \cos\alpha\sin\beta)$ 3sinalphacosbeta + 3cosalphasinbeta = 2sinalphacosbeta - 2cosalphasinbeta$3\sin\alpha\cos\beta + 3\cos\alpha\sin\beta = 2\sin\alpha\cos\beta - 2\cos\alpha\sin\beta$ ### Step 1: Rearranging Terms Grouping like terms together: 3sinalphacosbeta - 2sinalphacosbeta = -2cosalphasinbeta - 3cosalphasinbeta$3\sin\alpha\cos\beta - 2\sin\alpha\cos\beta = -2\cos\alpha\sin\beta - 3\cos\alpha\sin\beta$ sinalphacosbeta = -5cosalphasinbeta$\sin\alpha\cos\beta = -5\cos\alpha\sin\beta$ Dividing both sides by cosalphacosbeta$\cos\alpha\cos\beta$: fracsinalphacosalpha = -5fracsinbetacosbeta$\frac{\sin\alpha}{\cos\alpha} = -5\frac{\sin\beta}{\cos\beta}$ tanalpha = -5tanbeta$\tan\alpha = -5\tan\beta$ ### Step 2: Conclusion Comparing with the given equation tanalpha = ktanbeta$\tan\alpha = k\tan\beta$, we get k = -5$k = -5$. *Note by our answer (Bonus)*: Since alpha, beta in (0, fracpi2)$\alpha, \beta \in (0, \frac{\pi}{2})$, both tanalpha$\tan\alpha$ and tanbeta$\tan\beta$ must be positive. Hence, tanalpha = -5tanbeta$\tan\alpha = -5\tan\beta$ is not possible. The data is inconsistent, but the NTA key marks option (2) as correct. ### Pattern Recognition Standard expansion of sin(Apm B)$\sin(A\pm B)$ and grouping identical products to isolate tan(A)$\tan(A)$ and tan(B)$\tan(B)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Trigonometric Functions