If fraccos^248^circ-sin^212^circsin^224^circ-sin^26^circ=fracalpha+betasqrt52$\frac{\cos^{2}48^{\circ}-\sin^{2}12^{\circ}}{\sin^{2}24^{\circ}-\sin^{2}6^{\circ}}=\frac{\alpha+\beta\sqrt{5}}{2}$, where alpha,betain mathbbN$\alpha,\beta\in \mathbb{N}$, then alpha+beta$\alpha+\beta$ is equal to ____.

### Related Formula sin(A - B) = sin A cos B - cos A sin B$$\sin(A - B) = \sin A \cos B - \cos A \sin B$$ sin 2A = 2 sin A cos A$$\sin 2A = 2 \sin A \cos A$$ ### Core Logic csc 10^circ - sqrt3 sec 10^circ = frac1sin 10^circ - fracsqrt3cos 10^circ$$\csc 10^{\circ} - \sqrt{3} \sec 10^{\circ} = \frac{1}{\sin 10^{\circ}} - \frac{\sqrt{3}}{\cos 10^{\circ}}$$ = fraccos 10^circ - sqrt3sin 10^circsin 10^circ cos 10^circ$$= \frac{\cos 10^{\circ} - \sqrt{3}\sin 10^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}$$ ### Step 1: Sine Transformation Multiply and divide the numerator by 2 to inject standard trig values: = frac2 left(frac12cos 10^circ - fracsqrt32sin 10^circright)sin 10^circ cos 10^circ$$= \frac{2 \left(\frac{1}{2}\cos 10^{\circ} - \frac{\sqrt{3}}{2}\sin 10^{\circ}\right)}{\sin 10^{\circ} \cos 10^{\circ}}$$ Substitute sin 30^circ = 1/2$\sin 30^{\circ} = 1/2$ and cos 30^circ = sqrt3/2$\cos 30^{\circ} = \sqrt{3}/2$: = frac2 (sin 30^circ cos 10^circ - cos 30^circ sin 10^circ)sin 10^circ cos 10^circ$$= \frac{2 (\sin 30^{\circ} \cos 10^{\circ} - \cos 30^{\circ} \sin 10^{\circ})}{\sin 10^{\circ} \cos 10^{\circ}}$$ ### Step 2: Apply Multiple Angle identities The numerator becomes 2 sin(30^circ - 10^circ) = 2 sin 20^circ$2 \sin(30^{\circ} - 10^{\circ}) = 2 \sin 20^{\circ}$. Multiply and divide the denominator by 2 to construct sin 2theta$\sin 2\theta$: = frac4 sin 20^circ2 sin 10^circ cos 10^circ$$= \frac{4 \sin 20^{\circ}}{2 \sin 10^{\circ} \cos 10^{\circ}}$$ = frac4 sin 20^circsin 20^circ$$= \frac{4 \sin 20^{\circ}}{\sin 20^{\circ}}$$ = 4$= 4$ ### Pattern Recognition Any expression structured as A csc theta - B sec theta$A \csc \theta - B \sec \theta$ instantly signals a fraction collapse to 2 sin(alpha - theta) / sin(2theta)$2 \sin(\alpha - \theta) / \sin(2\theta)$ via multiplication by 2. When constants are 1$1$ and sqrt3$\sqrt{3}$, the anchor is always 30^circ$30^{\circ}$ or 60^circ$60^{\circ}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Trigonometric Functions

### Related Formula Tangent sum formula: tan 2alpha = tan[(alpha + beta) + (alpha - beta)] = fractan(alpha + beta) + tan(alpha - beta)1 - tan(alpha + beta)tan(alpha - beta)$$\tan 2\alpha = \tan[(\alpha + \beta) + (\alpha - \beta)] = \frac{\tan(\alpha + \beta) + \tan(\alpha - \beta)}{1 - \tan(\alpha + \beta)\tan(\alpha - \beta)}$$ ### Core Logic Given cos(alpha + beta) = -frac110 implies tan(alpha + beta) = -sqrt99 = -3sqrt11$\cos(\alpha + \beta) = -\frac{1}{10} \implies \tan(\alpha + \beta) = -\sqrt{99} = -3\sqrt{11}$. Given sin(alpha - beta) = frac38 implies tan(alpha - beta) = frac3sqrt55 = frac3sqrt5sqrt11$\sin(\alpha - \beta) = \frac{3}{8} \implies \tan(\alpha - \beta) = \frac{3}{\sqrt{55}} = \frac{3}{\sqrt{5}\sqrt{11}}$. ### Step 1: Simplify Tangent of 2 Alpha tan 2alpha = frac-3sqrt11 + frac3sqrt5sqrt111 + 3sqrt11 cdot frac3sqrt5sqrt11 = fracfrac-3sqrt55 + 3sqrt55fracsqrt5 + 9sqrt5 = frac3(1 - 11sqrt5)sqrt11(9 + sqrt5)$$\tan 2\alpha = \frac{-3\sqrt{11} + \frac{3}{\sqrt{5}\sqrt{11}}}{1 + 3\sqrt{11} \cdot \frac{3}{\sqrt{5}\sqrt{11}}} = \frac{\frac{-3\sqrt{55} + 3}{\sqrt{55}}}{\frac{\sqrt{5} + 9}{\sqrt{5}}} = \frac{3(1 - 11\sqrt{5})}{\sqrt{11}(9 + \sqrt{5})}$$ ### Step 2: Compare Constants Comparing with frac3(1 - rsqrt5)sqrt11(s + sqrt5)$\frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}$: r = 11, quad s = 9 implies r + s = 20$$r = 11, \quad s = 9 \implies r + s = 20$$ ### Pattern Recognition Express 2alpha$2\alpha$ as (alpha+beta) + (alpha-beta)$(\alpha+\beta) + (\alpha-\beta)$ to apply standard compound tangent formula directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Trigonometry

### Related Formula textQuadratic formula: y = frac-b pm sqrtb^2 - 4ac2a$$\text{Quadratic formula: } y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ ### Core Logic This is a quadratic equation in terms of csc theta$\csc \theta$. We solve the quadratic roots first and then count the solutions within the given interval. ### Step 1: Solve the quadratic equation Let y = csc theta$y = \csc \theta$: sqrt3 y^2 - 2(sqrt3 - 1)y - 4 = 0$$\sqrt{3} y^2 - 2(\sqrt{3} - 1)y - 4 = 0$$ y = frac2(sqrt3-1) pm sqrt4(sqrt3-1)^2 - 4(sqrt3)(-4)2sqrt3$$y = \frac{2(\sqrt{3}-1) \pm \sqrt{4(\sqrt{3}-1)^2 - 4(\sqrt{3})(-4)}}{2\sqrt{3}}$$ y = frac2(sqrt3-1) pm sqrt4(3 + 1 - 2sqrt3) + 16sqrt32sqrt3$$y = \frac{2(\sqrt{3}-1) \pm \sqrt{4(3 + 1 - 2\sqrt{3}) + 16\sqrt{3}}}{2\sqrt{3}}$$ y = frac2(sqrt3-1) pm sqrt16 + 8sqrt32sqrt3$$y = \frac{2(\sqrt{3}-1) \pm \sqrt{16 + 8\sqrt{3}}}{2\sqrt{3}}$$ Since 16 + 8sqrt3 = (2 + 2sqrt3)^2$16 + 8\sqrt{3} = (2 + 2\sqrt{3})^2$: y = frac2(sqrt3-1) pm (2 + 2sqrt3)2sqrt3$$y = \frac{2(\sqrt{3}-1) \pm (2 + 2\sqrt{3})}{2\sqrt{3}}$$ - Case 1 (+$+$ sign): y = frac4sqrt32sqrt3 = 2 implies sin theta = frac12$$y = \frac{4\sqrt{3}}{2\sqrt{3}} = 2 \implies \sin \theta = \frac{1}{2}$$ - Case 2 (-$-$ sign): y = frac-42sqrt3 = -frac2sqrt3 implies sin theta = -fracsqrt32$$y = \frac{-4}{2\sqrt{3}} = -\frac{2}{\sqrt{3}} \implies \sin \theta = -\frac{\sqrt{3}}{2}$$ ### Step 2: Count solutions in the interval Our interval is theta in left[-frac7pi6, frac4pi3 ight]$\theta \in \left[-\frac{7\pi}{6}, \frac{4\pi}{3} ight]$: - For sin theta = frac12$\sin \theta = \frac{1}{2}$: The general solutions are theta = fracpi6, frac5pi6$\theta = \frac{\pi}{6}, \frac{5\pi}{6}$. Within our interval, we have: theta = -frac7pi6, \, fracpi6, \, frac5pi6 quad (3 text solutions)$$\theta = -\frac{7\pi}{6}, \, \frac{\pi}{6}, \, \frac{5\pi}{6} \quad (3 \text{ solutions})$$ - For sin theta = -fracsqrt32$\sin \theta = -\frac{\sqrt{3}}{2}$: The general solutions are theta = -fracpi3, -frac2pi3$\theta = -\frac{\pi}{3}, -\frac{2\pi}{3}$. Within our interval, we have: theta = -frac2pi3, \, -fracpi3, \, frac4pi3 quad (3 text solutions)$$\theta = -\frac{2\pi}{3}, \, -\frac{\pi}{3}, \, \frac{4\pi}{3} \quad (3 \text{ solutions})$$ Summing the valid solutions: textTotal solutions = 3 + 3 = 6$$\text{Total solutions} = 3 + 3 = 6$$ ### Pattern Recognition Perfect Square discriminant: In JEE quadratics with irrational coefficients, the discriminant b^2-4ac$b^2-4ac$ almost always simplifies to a perfect square of the form (p + qsqrtr)^2$(p + q\sqrt{r})^2$. Double check your algebraic expansions if it doesn't simplify cleanly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometric Functions

### Related Formula Factorization of quadratic equations by splitting the middle term. ### Core Logic Treat the given equation as a standard quadratic in terms of costheta$\cos\theta$ and solve for its roots. ### Step 1: Factorization Split the middle term: 2sqrt2cos^2theta + 2costheta - sqrt6costheta - sqrt3 = 0$$2\sqrt{2}\cos^2\theta + 2\cos\theta - \sqrt{6}\cos\theta - \sqrt{3} = 0$$ 2costheta(sqrt2costheta + 1) - sqrt3(sqrt2costheta + 1) = 0$$2\cos\theta(\sqrt{2}\cos\theta + 1) - \sqrt{3}(\sqrt{2}\cos\theta + 1) = 0$$ (2costheta - sqrt3)(sqrt2costheta + 1) = 0$$(2\cos\theta - \sqrt{3})(\sqrt{2}\cos\theta + 1) = 0$$ ### Step 2: Solve for roots This yields two possible cases: costheta = fracsqrt32 quad textor quad costheta = -frac1sqrt2$$\cos\theta = \frac{\sqrt{3}}{2} \quad \text{or} \quad \cos\theta = -\frac{1}{\sqrt{2}}$$ ### Step 3: Count Solutions in Interval The given interval is [-2pi, 2pi]$[-2\pi, 2\pi]$, which covers two complete cycles of the cosine wave. * For costheta = fracsqrt32$\cos\theta = \frac{\sqrt{3}}{2}$, there are 2$2$ solutions per cycle implies 2 times 2 = 4$\implies 2 \times 2 = 4$ solutions. * For costheta = -frac1sqrt2$\cos\theta = -\frac{1}{\sqrt{2}}$, there are 2$2$ solutions per cycle implies 2 times 2 = 4$\implies 2 \times 2 = 4$ solutions. textTotal Solutions = 4 + 4 = 8$$\text{Total Solutions} = 4 + 4 = 8$$ ### Pattern Recognition Since both fracsqrt32$\frac{\sqrt{3}}{2}$ and -frac1sqrt2$-\frac{1}{\sqrt{2}}$ lie strictly between -1$-1$ and 1$1$, each horizontal line cuts the cosine function exactly twice per period (2pi$2\pi$). Across an interval of width 4pi$4\pi$, each value must yield exactly 4$4$ solutions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometric Functions