Question

$if \: \frac{x}{y} = \frac{cosA}{cosB} \: then \: \frac{xTanA + xTanB}{x + y} is \: equal \: to \: \: \ \\ \sf◆ \: tan \bigg( \frac{a + b}{2} \bigg) \\ \\ \sf◆Cot (A+B)\\ \\ \sf◆Tan(A+B) \\ \\ \sf◆Cot \frac {A+B}{2} \\ \\ ◆tan(2a+b​) \\ \\ ◆Cot(A+B) \\ \\ ◆Tan(A+B) \\ \\ ◆Cot2A+B​​$



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Answer 1

Answer:

$\tan( \frac{a + b}{2} )$

$\frac{x \tan(a) + y \tan(b) }{x + y} \: \: \: \: \: \: \:(1)$

$\frac{y \cos(a) }{ \cos(b) } \times \frac{ \sin(a) }{ \cos(a) } </p><p> + \frac{y \sin(b) }{ \cos(b) } \\ frac{y \cos(a) }{ \cos(b) } + y \\ \frac{y \sin(a) }{ \cos(b) } + \frac{y \sin(b) }{ \cos(b) } \\ \frac{y \cos(a) + y \cos(b) }{ \cos(b) } \\ = \frac{ \sin(a) + \sin(b) }{ \cos(a) + \cos(b) }$

$\sin(a) + \sin(b) = 2 \sin( \frac{a + b}{2} ) . \cos( \frac{a - b}{2} ) \\ by \\ \cos(a) + \cos(b) = 2 \cos( \frac{a + b}{2} ) . \cos( \frac{a - b}{2} ) \\ = \tan( \frac{a + b}{2} )$

Answer 2

Answer:

if

y

x

=

cosB

cosA

then

x+y

xTanA+xTanB

isequalto

◆tan(

2

a+b

)

◆Cot(A+B)

◆Tan(A+B)

◆Cot

2

A+B

◆tan(2a+b)

◆Cot(A+B)

◆Tan(A+B)

◆Cot2A+B

Step-by-step explanation:

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