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In any triangle,
$A B C=\cos ^2 \frac{A}{2}+\cos ^2 \frac{B}{2}+\cos ^2 \frac{C}{2}=$
Options:
$A B C=\cos ^2 \frac{A}{2}+\cos ^2 \frac{B}{2}+\cos ^2 \frac{C}{2}=$
Solution:
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Verified Answer
The correct answer is:
$2\left(1+\frac{r}{4 R}\right)$
We have,
$\begin{aligned} & \cos ^2 \frac{A}{2}+\cos ^2 \frac{B}{2}+\cos ^2 \frac{C}{2} \\ = & \frac{1}{2}\{(1+\cos A)+(1+\cos B)+(1+\cos C)\} \\ = & \frac{1}{2}\{3+\cos A+\cos B+\cos C\} \\ = & \frac{1}{2}\left\{3+1+\frac{r}{R}\right\}\left[\because \cos A+\cos B+\cos C=1+\frac{r}{R}\right] \\ = & \frac{1}{2}\left(4+\frac{r}{R}\right)=2+\frac{r}{2 R}\end{aligned}$
$\begin{aligned} & \cos ^2 \frac{A}{2}+\cos ^2 \frac{B}{2}+\cos ^2 \frac{C}{2} \\ = & \frac{1}{2}\{(1+\cos A)+(1+\cos B)+(1+\cos C)\} \\ = & \frac{1}{2}\{3+\cos A+\cos B+\cos C\} \\ = & \frac{1}{2}\left\{3+1+\frac{r}{R}\right\}\left[\because \cos A+\cos B+\cos C=1+\frac{r}{R}\right] \\ = & \frac{1}{2}\left(4+\frac{r}{R}\right)=2+\frac{r}{2 R}\end{aligned}$
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