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

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

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 $\frac{x \tan A-y \tan B}{x+y}=$

Similar Questions

Prove that $\frac{\sin x+\sin 3x}{\cos x+\cos 3x} = \tan 2x$.

If $\sin x = \frac{3}{5}$ and $\cos y = -\frac{12}{13}$,where $x$ and $y$ both lie in the second quadrant,find the value of $\sin (x+y)$.

If $\tan A = \frac{5}{6}$ and $\tan B = \frac{1}{11}$,then $A + B = $

$(\cos \alpha + \cos \beta )^2 + (\sin \alpha + \sin \beta )^2 = $

If $\cos (\alpha + \beta) = \frac{3}{5}$,$\sin (\alpha - \beta) = \frac{5}{13}$ and $0 < \alpha, \beta < \frac{\pi}{4}$,then $\tan (2\alpha)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module