यदि tan A = frac{1}{2} और tan B = frac{1}{3} | vedclass

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(B) दिया गया है $	an A = \frac{1}{2}$ और $	an B = \frac{1}{3}$.हम जानते हैं कि $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B} = \frac{\fra

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$\sin 2B$

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(B) दिया गया है $	an A = \frac{1}{2}$ और $	an B = \frac{1}{3}$.हम जानते हैं कि $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B} = \frac{\frac{1}{2} + \frac{1}{3}}{1 - (\frac{1}{2} 	imes \frac{1}{3})} = \frac{\frac{5}{6}}{1 - \frac{1}{6}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1$.चूंकि $	an(A + B) = 1$,इसलिए $A + B = 45^\circ$.अतः,$2A = 90^\circ - 2B$.दोनों पक्षों का कोसाइन लेने पर,$\cos 2A = \cos(90^\circ - 2B) = \sin 2B$.

यदि $\tan A = \frac{1}{2}$ और $\tan B = \frac{1}{3}$ है,तो $\cos 2A = $

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