Let alpha and beta be real numbers such that | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $\sin(\alpha+\beta) = \frac{1}{3}$ and $\cos(\alpha-\beta) = \frac{2}{3}$.Let $E = \left(\frac{\sin \alpha}{\cos \beta} + \frac{\cos \beta}{

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given $\sin(\alpha+\beta) = \frac{1}{3}$ and $\cos(\alpha-\beta) = \frac{2}{3}$.Let $E = \left(\frac{\sin \alpha}{\cos \beta} + \frac{\cos \beta}{\sin \alpha} + \frac{\cos \alpha}{\sin \beta} + \frac{\sin \beta}{\cos \alpha}\right)^2$.Grouping terms: $E = \left(\frac{\sin \alpha \sin \beta + \cos \alpha \cos \beta}{\cos \beta \sin \beta} + \frac{\cos \alpha \cos \beta + \sin \alpha \sin \beta}{\sin \alpha \cos \alpha}\right)^2$.Using $\cos(\alpha-\beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$,we get:$E = \cos^2(\alpha-\beta) \left(\frac{1}{\cos \beta \sin \beta} + \frac{1}{\sin \alpha \cos \alpha}\right)^2$.$E = \cos^2(\alpha-\beta) \left(\frac{2}{\sin 2\beta} + \frac{2}{\sin 2\alpha}\right)^2 = 4 \cos^2(\alpha-\beta) \left(\frac{\sin 2\alpha + \sin 2\beta}{\sin 2\alpha \sin 2\beta}\right)^2$.Using $\sin 2\alpha + \sin 2\beta = 2 \sin(\alpha+\beta) \cos(\alpha-\beta)$ and $\sin 2\alpha \sin 2\beta = \frac{1}{2}(\cos(2\alpha-2\beta) - \cos(2\alpha+2\beta)) = \cos^2(\alpha-\beta) - \sin^2(\alpha+\beta)$:$E = 4 \cos^2(\alpha-\beta) \left(\frac{2 \sin(\alpha+\beta) \cos(\alpha-\beta)}{\cos^2(\alpha-\beta) - \sin^2(\alpha+\beta)}\right)^2$.Substituting values: $\sin(\alpha+\beta) = \frac{1}{3}$,$\cos(\alpha-\beta) = \frac{2}{3}$.$E = 4 \left(\frac{4}{9}\right) \left(\frac{2 \cdot \frac{1}{3} \cdot \frac{2}{3}}{\frac{4}{9} - \frac{1}{9}}\right)^2 = \frac{16}{9} \left(\frac{4/9}{3/9}\right)^2 = \frac{16}{9} \left(\frac{4}{3}\right)^2 = \frac{16}{9} \cdot \frac{16}{9} = \frac{256}{81} \approx 3.16$.Wait,re-evaluating the expression: $\left(\frac{\sin \alpha}{\cos \beta} + \frac{\cos \beta}{\sin \alpha}\right) + \left(\frac{\cos \alpha}{\sin \beta} + \frac{\sin \beta}{\cos \alpha}\right) = \frac{\sin^2 \alpha + \cos^2 \beta}{\sin \alpha \cos \beta} + \frac{\cos^2 \alpha + \sin^2 \beta}{\cos \alpha \sin \beta}$.Actually,the expression simplifies to $\frac{\cos(\alpha-\beta)}{\sin \alpha \cos \beta} + \frac{\cos(\alpha-\beta)}{\cos \alpha \sin \beta} = \cos(\alpha-\beta) \frac{\sin(\alpha+\beta)}{\sin \alpha \cos \alpha \sin \beta \cos \beta} = \frac{4 \cos(\alpha-\beta) \sin(\alpha+\beta)}{\sin 2\alpha \sin 2\beta}$.Using $\sin 2\alpha \sin 2\beta = \cos(2\alpha-2\beta) - \cos(2\alpha+2\beta) = 2\cos^2(\alpha-\beta) - 1 - (1 - 2\sin^2(\alpha+\beta)) = 2(\frac{4}{9}) - 2 + 2(\frac{1}{9}) = \frac{8}{9} - 2 + \frac{2}{9} = -\frac{8}{9}$.$E = \left(\frac{4 \cdot (2/3) \cdot (1/3)}{-8/9}\right)^2 = \left(\frac{8/9}{-8/9}\right)^2 = (-1)^2 = 1$.The greatest integer is $1$.

Similar Questions

The value of $\frac{1}{4} \tan \frac{\pi}{8} + \frac{1}{8} \tan \frac{\pi}{16} + \frac{1}{16} \tan \frac{\pi}{32} + \dots \infty$ terms is equal to-

The expression $2 \cos \frac{\pi}{13} \cos \frac{9\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13}$ is equal to

If $x=\frac{\sin^3 \theta}{\cos^2 \theta}$ and $y=\frac{\cos^3 \theta}{\sin^2 \theta}$,where $\sin \theta+\cos \theta=\frac{1}{2}$,then $x+y=$

If $\cos \left(\frac{\alpha-\beta}{2}\right)=2 \cos \left(\frac{\alpha+\beta}{2}\right)$,then $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}=$

$\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module