Question

x/y=cosA/cosB,then xtanA+ytanB/x+y​

Answer 1

Answer:

ANSWER.

$\frac{x \tan(a) + y \tan(b) }{x + y} = \frac{ \sin(a) + \sin(b) }{ \cos(a)+\cos(b) } \\ \\ ...(answer)$

Step-by-step explanation:

GIVEN :

$\frac{x}{y} = \frac{ \cos(a) }{ \cos(b) }$

TO FIND :

$value \: of \: \: \frac{x \tan(a) + y \tan(b) }{x + y} .$

SOLUTION :

$\frac{x}{y} = \frac{ \cos(a) }{ \cos(b) } \\ \\ \frac{x \sin(b) }{y \sin(a) } = \frac{ \cos(a) \sin(b) }{ \sin(a) \cos(b) } \\ \\ \frac{x \sin(b) }{y \sin(a) } = \frac{ \tan(b) }{ \tan(a) } \\ \\ \frac{x \tan(a) }{y \tan(b) } = \frac{ \sin(a) }{ \sin(b) } \\ \\ applying \: componendo \: \: we \: get \\ \\ \frac{x \tan(a) + y \tan(b) }{y \tan(b) } = \frac{ \sin(a) + \sin(b) }{ \sin(b) } \\ \\ ...(1) \\ \\ \\ \\ now \: \: \: \: \: \: \frac{x}{y} = \frac{ \cos(a) }{ \cos(b) } \\ \\applying \: componendo \\ \\ \frac{x + y}{y} = \frac{ \cos(a) + \cos(b) }{ \cos(b) } \\ \\ ...(2) \\ \\ from \: (1) \: and \: (2) \: we \: get \\ \\ \\ \frac{x \tan(a) + y \tan(b) }{x + y} = \tan(b) . \frac{ \frac{ \sin(a) + \sin(b) }{ \sin(b) } }{ \frac{ \cos(a)+\cos(b) }{ \cos(b) } } \\ \\ \frac{x \tan(a) + y \tan(b) }{x + y} = \frac{ \sin(a) + \sin(b) }{ \cos(a) +\cos(b)} \\ \\ ...(answer)$