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Determinant

In Δ ABC ...

Question

In $Δ$ $A B C$ the sides opposite to angles $A, B, C$ are denoted by $a, b, c$ respectively. Let $r_{1}$ and $r_{2}$ be the ex-radius of the traingle.
In a triangle $A B C, \frac{r_{1} + r_{2}}{1 + cos C}$ is equal to?

\frac{2 a b}{c Δ}

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\frac{a + b}{c Δ}

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\frac{a b c}{2 Δ}

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\frac{a b c}{2 Δ^{2}}

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Solution

The correct option is A $\frac{a b c}{2 Δ}$
The value of $\frac{r_{1} + r_{2}}{1 + cos C} = \frac{\frac{Δ}{s - a} + \frac{Δ}{s - b}}{1 + \frac{b^{2} + a^{2} - c^{2}}{2 a b}}$
$= \frac{Δ c}{(s - a) (s - b)} \cdot \frac{2 a b}{(a + b)^{2} - c^{2}}$
$= \frac{Δ c}{(s - a) (s - b)} \cdot \frac{2 a b}{2 s (2 s - 2 c)}$
$= \frac{a b c}{2 Δ}$
$∴ \frac{r_{1} + r_{2}}{1 + cos C} = \frac{a b c}{2 Δ}$

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In Δ ABC the sides opposite to angles A,B,C are denoted by a,b,c respectively. Let r1 and r2 be the ex-radius of the traingle.In a triangle ABC,r1+r21+cosC is equal to?

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