Question

if alpha, beta, gamma are the lengths of the altitudes of triangle ABC then one by Alpha Plus One by beta minus 1 by Gamma minus 2 a b by a + b + c of delta into cos square C by 2​

Answer 1

Answer:

The question is

$1/\alpha + 1/\beta - 1/r - \frac{2ab}{(a+b+c)delta} cos^{2} (\frac{C}{2} )$

So, We have to prove

$1/\alpha + 1/\beta - 1/r = \frac{2ab}{(a+b+c)delta} cos^{2} (\frac{C}{2} )$

Step-by-step explanation:

In order to prove above equation, We take R.H.S

$\frac{2ab}{(a+b+c)delta} cos^{2} (\frac{C}{2} )$

We use half angle formula here:

$cos^{2}\alpha = \frac{1+cos\alpha }{2}$

So, R.H.S become

$\frac{2ab}{(a+b+c)delta} (\frac{cosC+1}{2})$

$\frac{abcosC+ab}{(a+b+c)delta}$

Now, from $cosC = \frac{a^{2}+b^{2}-c^{2} }{2ab}$

We get $abcosC = \frac{a^{2}+b^{2}-c^{2} }{2}$

So, by putting the value of abcosC in R.H.S equation we get

$\frac{\frac{a^{2}+b^{2}-c^{2} }{2} + ab}{(a+b+c)delta}$

$\frac{{a^{2}+b^{2}-c^{2} } + 2ab}{2(a+b+c)delta}$

$\frac{{(a+b)^{2}-c^{2} }}{2(a+b+c)delta}$

So, $\frac{{(a+b+c) (a+b-c)}}{2(a+b+c)delta}$

$\frac{{(a+b-c)}}{2delta}$

We can write it as

$\frac{a}{2delta} + \frac{b}{2delta} - \frac{c}{2delta}$

delta  = $\frac{1}{2}a(\alpha )=\frac{1}{2}b(\beta ) = \frac{1}{2}c (r)$

So, R.H.S become

$\frac{1}{\alpha } + \frac{1}{\beta} - \frac{1}{r}$ = L.H.S

Hence proved that R.H.S = L.H.S

I hope this answer may help you

Answer 2

Answer:

answer is option d ( delta )

Step-by-step explanation: