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If $P_1, P_2$ and $P_3$ are the lengths of the altitudes drawn from the vertices $A, B$ and $C$ of $\triangle A B C$ respectively, then $\frac{\cos A}{P_1}+\frac{\cos B}{P_2}+\frac{\cos C}{P_3}=$
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Verified Answer
The correct answer is:
$\frac{1}{\mathrm{R}}$
We know that:
$$
P_1=\frac{2 \Delta}{a}, P_2=\frac{2 \Delta}{b}, P_3=\frac{2 \Delta}{c}
$$
Now, $\frac{\cos A}{P_1}+\frac{\cos B}{P_2}+\frac{\cos C}{P_3}$
$$
\begin{aligned}
& =\frac{a \cos A}{2 \Delta}+\frac{b \cos B}{2 \Delta}+\frac{c \cos C}{2 \Delta} \\
& =\frac{2 R \sin A \cos A+2 R \sin B \cos B+2 R \sin C \cos C}{2 \Delta} \\
& =\frac{R(\sin 2 A+\sin 2 B+\sin 2 C)}{2 \Delta} \\
& =\frac{2 R(\sin (A+B) \cos (A-B)+\sin C \cos C)}{2 \Delta} \\
& =\frac{2 R(\sin C \cos (A-B)-\sin C \cos (A+B))}{2 \Delta} \\
& =\frac{4 R \sin C[\sin B \sin C]}{2 \Delta} \\
& =\frac{4 R\left(\frac{a}{2 R}\right)\left(\frac{b}{2 R}\right)\left(\frac{c}{2 R}\right)}{2 \Delta}=\frac{1}{R^2} \frac{a b c}{4 \Delta} \\
& =\frac{1}{R^2} \times R=\frac{1}{R}
\end{aligned}
$$
$$
P_1=\frac{2 \Delta}{a}, P_2=\frac{2 \Delta}{b}, P_3=\frac{2 \Delta}{c}
$$
Now, $\frac{\cos A}{P_1}+\frac{\cos B}{P_2}+\frac{\cos C}{P_3}$
$$
\begin{aligned}
& =\frac{a \cos A}{2 \Delta}+\frac{b \cos B}{2 \Delta}+\frac{c \cos C}{2 \Delta} \\
& =\frac{2 R \sin A \cos A+2 R \sin B \cos B+2 R \sin C \cos C}{2 \Delta} \\
& =\frac{R(\sin 2 A+\sin 2 B+\sin 2 C)}{2 \Delta} \\
& =\frac{2 R(\sin (A+B) \cos (A-B)+\sin C \cos C)}{2 \Delta} \\
& =\frac{2 R(\sin C \cos (A-B)-\sin C \cos (A+B))}{2 \Delta} \\
& =\frac{4 R \sin C[\sin B \sin C]}{2 \Delta} \\
& =\frac{4 R\left(\frac{a}{2 R}\right)\left(\frac{b}{2 R}\right)\left(\frac{c}{2 R}\right)}{2 \Delta}=\frac{1}{R^2} \frac{a b c}{4 \Delta} \\
& =\frac{1}{R^2} \times R=\frac{1}{R}
\end{aligned}
$$
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