Question

If $\alpha ,\beta ,\gamma$ are the lengths of the altitudes of $\Delta ABC$, then $\frac{1}{\alpha }+\frac{1}{\beta }-\frac{1}{\gamma }-\frac{2ab}{(a+b+c)\Delta }{\cos }^2\frac{C}{2}=$

• A. 0
• B. 1
• C. 2s
• D. $\Delta$

Answer 1

The correct option is A 0

Let us assume that Given Triangle is an Equilateral Triangle of side 1 unit means $a=b=c=1$

in which $\alpha =\beta =\gamma =1\times sin{60}^o=\frac{\sqrt{3}}{2}$

and $cosA=cosB=cosC=cos{60}^0=\frac{1}{2}$

As we know that Area of Equilateral Triangle $=△=\frac{\sqrt{3}}{4}\times 1^2$

Now, $\frac{1}{\alpha }+\frac{1}{\beta }-\frac{1}{\gamma }=\frac{2}{\sqrt{3}}$

and $\frac{2ab}{(a+b+c)△}co{s}^2\frac{C}{2}=\frac{2ab}{(a+b+c)△}\frac{1+cosC}{2}=\frac{2}{\sqrt{3}}$

And

$\frac{1}{\alpha }+\frac{1}{\beta }-\frac{1}{\gamma }-\frac{2ab}{(a+b+c)△}co{s}^2\frac{C}{2}=\frac{2}{\sqrt{3}}-\frac{2}{\sqrt{3}}=0$

therefore Answer is $A$