In Triangle ABC the value of r 1- r 2-1-cos C is always equal toA- 2 rB- 2 r2-RC- 2 RD- 2 R2-r

The correct option is C 2R r_1+r_2=4R [sin A 2.cos B 2.cos C 2+sin B 2.cos A 2.cos C 2 ] =4R [cos C 2 [sin A 2.cos B 2+sin B 2.cos A 2 ] ] =4R [cos^2 C 2 ]=2 ...

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Standard XII

Mathematics

Monotonically Increasing Functions

In Triangle A...

Question

In Triangle ABC the value of $\frac{r_{1} + r_{2}}{1 + c o s C}$ is always equal to

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Solution

The correct option is B
2R

$r_{1} + r_{2} = 4 R [s i n \frac{A}{2} . c o s \frac{B}{2} . c o s \frac{C}{2} + s i n \frac{B}{2} . c o s \frac{A}{2} . c o s \frac{C}{2}]$
= $4 R [c o s \frac{C}{2} [s i n \frac{A}{2} . c o s \frac{B}{2} + s i n \frac{B}{2} . c o s \frac{A}{2}]]$
= $4 R [c o s^{2} \frac{C}{2}] = 2 R (1 + c o s C)$
$\frac{r_{1} + r_{2}}{1 + c o s C} = 2 R$

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In Triangle ABC the value of r1+r21+cosC is always equal to

The correct option is B 2R r1+r2=4R[sin A2.cos B2.cos C2+sin B2.cos A2.cos C2] =4R[cos C2[sin A2.cos B2+sin B2.cos A2]] =4R[cos2 C2]=2R(1+cos C) r1+r21+cos C=2R

In Triangle ABC the value of $\frac{r_{1} + r_{2}}{1 + c o s C}$ is always equal to