Question

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

• A. tan A + tanB + tan C
• B. cot A + cot B + cot C
• C. cos A + cos B +cos C
• D. cosec A + cosec B + cosec C

Answer 1

The correct option is A tan A + tanB + tan C

$RF=z,QF=y,PD=x$

$AP$ and $BC$ are intersecting chords.

$AD\cdot DP=BD\cdot DC$

$\Rightarrow x=\frac{BD\cdot DC}{AD}$

$\frac{a}{2x}=\frac{(BD+DC)AD}{2BD\cdot DC}$

$=\frac{1}{2}(\frac{AD}{BD}+\frac{AD}{DC})$

$=\frac{1}{2}(\tan B+\tan C)$

Similarly, $\frac{b}{2x}=\frac{1}{2}(\tan C+\tan A)$

$\frac{c}{x}=\frac{1}{2}(\tan A+\tan B)$

$\therefore \frac{a}{2x}+\frac{b}{2x}+\frac{c}{2x}=\tan A+\tan B+\tan C$