If two acute angles A and B are such that A n | vedclass

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Question

(A) Given: $\frac{x}{y} = \frac{\cos A}{\cos B}$ Divide the numerator and denominator of the expression by $y$: $\frac{x 	an A - y 	an B}{x + y} = \

Vedclass · Accepted Answer

$	an \left(\frac{A-B}{2}ight)$

Vedclass · Answer

(A) Given: $\frac{x}{y} = \frac{\cos A}{\cos B}$ Divide the numerator and denominator of the expression by $y$: $\frac{x 	an A - y 	an B}{x + y} = \frac{\frac{x}{y} 	an A - 	an B}{\frac{x}{y} + 1}$ Substitute $\frac{x}{y} = \frac{\cos A}{\cos B}$: $= \frac{\frac{\cos A}{\cos B} 	an A - 	an B}{\frac{\cos A}{\cos B} + 1} = \frac{\frac{\cos A \sin A}{\cos B \cos A} - 	an B}{\frac{\cos A + \cos B}{\cos B}} = \frac{\frac{\sin A}{\cos B} - \frac{\sin B}{\cos B}}{\frac{\cos A + \cos B}{\cos B}}$ $= \frac{\sin A - \sin B}{\cos A + \cos B}$ Using sum-to-product formulas: $= \frac{2 \sin \left(\frac{A-B}{2}ight) \cos \left(\frac{A+B}{2}ight)}{2 \cos \left(\frac{A+B}{2}ight) \cos \left(\frac{A-B}{2}ight)}$ $= \frac{\sin \left(\frac{A-B}{2}ight)}{\cos \left(\frac{A-B}{2}ight)} = 	an \left(\frac{A-B}{2}ight)$

If two acute angles $A$ and $B$ are such that $A \neq B$ and $\frac{x}{y}=\frac{\cos A}{\cos B}$,then $\frac{x \tan A-y \tan B}{x+y}=$

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