Question

In triangleABC r1+r2 /1+cosC
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Answer 1

Answer:

Please see the attachment

Answer 2

r1+r2 /1+cosC = abc/2Δ

To Find:

The value of r1+r2 / 1+cosC

Solution:

In ΔABC

we know that,

$r_{1} = 4Rsin\frac{A}{2}cos\frac{B}{2} cos\frac{C}{2} \\\\r_{2} = 4Rsin\frac{B}{2}cos\frac{A}{2} cos\frac{C}{2}$

Let t =  r1+r2 / 1+cosC

Also, 1+cosC= cos^2(C/2)

$t = \frac{4Rsin\frac{A}{2}cos\frac{B}{2} cos\frac{C}{2}+4Rsin\frac{B}{2}cos\frac{A}{2} cos\frac{C}{2}}{cos^{2} \frac{C}{2} }$

$t = \frac{4Rsin\frac{A}{2}cos\frac{B}{2} +4Rsin\frac{B}{2}cos\frac{A}{2} }{cos\frac{C}{2} } \\\\t = \frac{4Rsin\frac{A+B}{2} }{cos\frac{C}{2} } \\\\t= 4R$

t = abc/2Δ

Therefore,  r1+r2 /1+cosC = abc/2Δ

#SPJ3