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Question

If $$P_1,P_2,P_3$$ are the altitudes of a triangle from the vertices $$A,B,C$$ & $$\Delta $$ denotes the area of triangle, then $$\frac{1}{P_1}+\frac{1}{P_2}-\frac{1}{P_3}=\frac{2ab}{(a+b+c){\Delta}}cos^2\frac{C}{2}$$.


A
True
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B
False
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Solution

The correct option is A True
Since $$\Delta=\dfrac{1}{2}a{p}_{1}\Rightarrow\,\dfrac{1}{{p}_{1}}=\dfrac{a}{2\Delta}$$

$$\Delta=\dfrac{1}{2}b{p}_{2}\Rightarrow\,\dfrac{1}{{p}_{2}}=\dfrac{b}{2\Delta}$$

$$\Delta=\dfrac{1}{2}c{p}_{3}\Rightarrow\,\dfrac{1}{{p}_{3}}=\dfrac{c}{2\Delta}$$

$$\therefore\,\dfrac{1}{{p}_{1}}+\dfrac{1}{{p}_{2}}-\dfrac{1}{{p}_{3}}=\dfrac{1}{2\Delta}\left(a+b-c\right)$$

$$=\dfrac{2\left(s-c\right)}{2\Delta}=\dfrac{s-c}{\Delta}=\dfrac{s\left(s-c\right)}{ab}.\dfrac{ab}{s\Delta}$$

$$=\dfrac{ab}{\left(\dfrac{a+b+c}{2}\right)\Delta}{\cos}^{2}{\dfrac{C}{2}}$$

$$=\dfrac{2ab}{\left(a+b+c\right)\Delta}{\cos}^{2}{\dfrac{C}{2}}$$
Hence the given statement is true.

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