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: $(\cos x+\cos y)^{2}+(\sin x-\sin y)^{2}=4 \cos ^{2} \frac{x+y}{2}$

यदि $(1+\tan \alpha)(1+\tan 4 \alpha)=2$ और $\alpha \in \left(0, \frac{\pi}{16}\right)$ है,तो $\alpha$ का मान ज्ञात कीजिए।

$1 + \cos 2x + \cos 4x + \cos 6x = $

$\frac{\tan 52^{\circ} - \tan 38^{\circ}}{\tan 14^{\circ}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module