यदि cos(alpha + beta) = frac{4}{5} और sin(alp | vedclass

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

Question

(B) दिया है $\cos(\alpha + \beta) = \frac{4}{5}$. चूँकि $0 \le \alpha, \beta \le \frac{\pi}{4}$,इसलिए $0 \le \alpha + \beta \le \frac{\pi}{2}$,अतः $

Vedclass · Accepted Answer

$\frac{56}{33}$

Vedclass · Answer

(B) दिया है $\cos(\alpha + \beta) = \frac{4}{5}$. चूँकि $0 \le \alpha, \beta \le \frac{\pi}{4}$,इसलिए $0 \le \alpha + \beta \le \frac{\pi}{2}$,अतः $	an(\alpha + \beta) = \frac{3}{4}$.दिया है $\sin(\alpha - \beta) = \frac{5}{13}$. अतः $	an(\alpha - \beta) = \frac{5}{12}$.अब,$	an 2\alpha = 	an((\alpha + \beta) + (\alpha - \beta))$.$	an 2\alpha = \frac{\frac{3}{4} + \frac{5}{12}}{1 - (\frac{3}{4} 	imes \frac{5}{12})} = \frac{14/12}{33/48} = \frac{56}{33}$.

यदि $\cos(\alpha + \beta) = \frac{4}{5}$ और $\sin(\alpha - \beta) = \frac{5}{13}$,जहाँ $0 \le \alpha, \beta \le \frac{\pi}{4}$ है,तो $\tan 2\alpha = $

Similar Questions

सिद्ध कीजिए कि $\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x$.

सिद्ध कीजिए कि $\frac{\sin (x+y)}{\sin (x-y)} = \frac{\tan x + \tan y}{\tan x - \tan y}$.

यदि $\frac{1-\tan \theta}{1+\tan \theta}=\frac{1}{\sqrt{3}}$,जहाँ $\theta \in \left(0, \frac{\pi}{2}\right)$,तो $\theta=$

यदि $m \cos (\alpha+\beta)-n \cos (\alpha-\beta)=m \cos (\alpha-\beta)+n \cos (\alpha+\beta)$ है,तो $\tan \alpha \tan \beta=$

$\cos \alpha \sin (\beta-\gamma) + \cos \beta \sin (\gamma-\alpha) + \cos \gamma \sin (\alpha-\beta)$ का मान क्या है?

Vedclass Test Series

Exam Paper Generator

Online Exam Module