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Q.

As shown in the figure ,AD is the altitude on BC and AD produced meets the circumcircle of Triangle ABC at P where DP = x. Similarly, EQ = y and FR = z. If a, b, c respectively denotes the sides BC, CA and AB, then $\frac{a}{2 x} + \frac{b}{2 y} + \frac{c}{2 z}$ has the value equal to

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a

$\cot A + \cos B + \cot C$

b

$cosec A + cosec B + cosec C$

c

$\cos A + \cos B + \cos C$

d

$\tan A + \tan B + \tan C$

answer is A.

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Detailed Solution

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$\begin{array}{l} BD = xtan C in △ PDB \\ and DC = xtan B for △ PDC \\ ∴ BD + DC = a = x (\tan B + \tan C) \\ \frac{a}{x} = \tan B + \tan C \\ Similarly, \frac{b}{y} = \tan A + \tan C \\ \frac{c}{z} = \tan A + \tan B \end{array}$
$∴ \frac{a}{2 x} + \frac{b}{2 y} + \frac{c}{2 z} = \frac{1}{2} (\frac{a}{x} + \frac{b}{y} + \frac{c}{z}) \tan A + \tan B + \tan C$

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