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

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Question

(B) दिया गया है $\cos (\alpha+\beta)=\frac{4}{5}$। चूँकि $0 \leq \alpha, \beta \leq \frac{\pi}{4}$,हमारे पास $0 \leq \alpha+\beta \leq \frac{\pi}{2}$

Vedclass · Accepted Answer

$\frac{56}{33}$

Vedclass · Answer

(B) दिया गया है $\cos (\alpha+\beta)=\frac{4}{5}$। चूँकि $0 \leq \alpha, \beta \leq \frac{\pi}{4}$,हमारे पास $0 \leq \alpha+\beta \leq \frac{\pi}{2}$ है,इसलिए $	an (\alpha+\beta)=\frac{3}{4}$।दिया गया है $\sin (\alpha-\beta)=\frac{5}{13}$। चूँकि $0 \leq \alpha, \beta \leq \frac{\pi}{4}$,हमारे पास $-\frac{\pi}{4} \leq \alpha-\beta \leq \frac{\pi}{4}$ है,इसलिए $	an (\alpha-\beta)=\frac{5}{12}$।हम जानते हैं कि $2\alpha = (\alpha+\beta)+(\alpha-\beta)$।अतः,$	an 2 \alpha = 	an \{(\alpha+\beta)+(\alpha-\beta)\} = \frac{	an (\alpha+\beta)+	an (\alpha-\beta)}{1-	an (\alpha+\beta) \cdot 	an (\alpha-\beta)}$।मान रखने पर: $	an 2 \alpha = \frac{\frac{3}{4}+\frac{5}{12}}{1-(\frac{3}{4} \cdot \frac{5}{12})} = \frac{\frac{14}{12}}{\frac{33}{48}} = \frac{56}{33}$।

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

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