ધારો કે cos(alpha+beta)=-frac{1}{10} અને sin( | vedclass

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

Question

(C) આપેલ છે $\cos(\alpha+\beta) = -\frac{1}{10}$. $0 < \alpha < \frac{\pi}{3}$ અને $0 < \beta < \frac{\pi}{4}$ હોવાથી, $0 < \alpha+\bet

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(C) આપેલ છે $\cos(\alpha+\beta) = -\frac{1}{10}$. $0 < \alpha < \frac{\pi}{3}$ અને $0 < \beta < \frac{\pi}{4}$ હોવાથી, $0 < \alpha+\beta < \frac{7\pi}{12}$. $\cos(\alpha+\beta) < 0$ હોવાથી, $\frac{\pi}{2} < \alpha+\beta < \frac{7\pi}{12}$.
તેથી, $\sin(\alpha+\beta) = \sqrt{1 - (-\frac{1}{10})^2} = \sqrt{\frac{99}{100}} = \frac{3\sqrt{11}}{10}$.
તેથી, $	an(\alpha+\beta) = \frac{3\sqrt{11}/10}{-1/10} = -3\sqrt{11}$.
આપેલ છે $\sin(\alpha-\beta) = \frac{3}{8}$. $0 < \alpha < \frac{\pi}{3}$ અને $0 < \beta < \frac{\pi}{4}$ હોવાથી, $-\frac{\pi}{4} < \alpha-\beta < \frac{\pi}{3}$. $\sin(\alpha-\beta) > 0$ હોવાથી, $0 < \alpha-\beta < \frac{\pi}{3}$.
તેથી, $\cos(\alpha-\beta) = \sqrt{1 - (\frac{3}{8})^2} = \sqrt{\frac{55}{64}} = \frac{\sqrt{55}}{8}$.
તેથી, $	an(\alpha-\beta) = \frac{3/8}{\sqrt{55}/8} = \frac{3}{\sqrt{55}}$.
હવે, $	an 2\alpha = 	an((\alpha+\beta)+(\alpha-\beta)) = \frac{	an(\alpha+\beta) + 	an(\alpha-\beta)}{1 - 	an(\alpha+\beta)	an(\alpha-\beta)}$.
કિંમતો મૂકતા, $	an 2\alpha = \frac{-3\sqrt{11} + \frac{3}{\sqrt{55}}}{1 - (-3\sqrt{11})(\frac{3}{\sqrt{55}})} = \frac{\frac{-3\sqrt{11}\sqrt{55} + 3}{\sqrt{55}}}{1 + \frac{9\sqrt{11}}{\sqrt{55}}} = \frac{-3(11\sqrt{5}) + 3}{\sqrt{55} + 9\sqrt{11}} = \frac{3(1 - 11\sqrt{5})}{\sqrt{11}(\sqrt{5} + 9)}$.
$\frac{3(1-r\sqrt{5})}{\sqrt{11}(s+\sqrt{5})}$ સાથે સરખાવતા, આપણને $r=11$ અને $s=9$ મળે છે.
તેથી, $r+s = 11+9 = 20$.

Similar Questions

જો $\cos \alpha + \cos \beta = \frac{1}{3}$ અને $\sin \alpha + \sin \beta = \frac{1}{4}$ હોય,તો $\cos (\alpha + \beta) = $

$\frac{3 + \cot 76^{\circ} \cot 16^{\circ}}{\cot 76^{\circ} + \cot 16^{\circ}}$ ની કિંમત શોધો.

જો $\frac{\sin(x + y)}{\sin(x - y)} = \frac{a + b}{a - b}$ હોય,તો $\frac{\tan x}{\tan y}$ ની કિંમત શોધો.

સાબિત કરો કે $\frac{\sin (x+y)}{\sin (x-y)} = \frac{\tan x + \tan y}{\tan x - \tan y}$.

$\tan 75^\circ - \cot 75^\circ = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module