Question

If a^2 b^2 c^2 are in ap then prove that cota cotb cotc are in ap

Answer 1

$a^2,b^2,c^2$ are in A.P. then prove that cotA, cotB, cotC are in A.P.

Step-by-step explanation:

We will show that if cotA, cotB and cotC are in AP then $a^2,b^2,c^2$ will also be in AP.

Since cotA, cotB ,cotC are in AP then

2cotB = cotA + cotC

Put the trigonometry  ratios, we get following equation

$2\frac{cosB}{sinB} = \frac{cosA}{sinA} + \frac{cosC}{sinC}$ -------------(1)

As we know, $\frac{SinA}{a} = \frac{SinB}{b} = \frac{SinC}{c} = k$ (constant, suppose)

Thus, sinA = ka, sinB = kb, sinC = kc

and  $cosA= \frac{b^2+c^2-a^2}{2bc}$ , $cosB= \frac{a^2+c^2-b^2}{2ac}$  and  $cosC= \frac{b^2+a^2-c^2}{2ab}$

Put all the above values in eq. (1)  we get,

$\frac{2}{kb}\frac{a^2+c^2-b^2}{2ac} = \frac{1}{ka}\frac{b^2+c^2-a^2}{2bc} + \\\frac{1}{kc}\frac{b^2+a^2-c^2}{2ab}$

$2a^2+2c^2-2b^2=b^2+c^2-a^2+b^2+a^2-c^2$

$2b^2=a^2+c^2$

Thus, $a^2,b^2,c^2$ are in A.P.

Answer 2

cotB - cotA=cotC - cotB cosB/sinB - cosA/sona=cosC/sinC-cosB/sinB , sin(A-B) sin(A+B) =sin(B-C) sinB+C),sin^2A-sin^2B=sin^2B-sin^2C,2sin^2B=sin^2A+sin^2C,2b^2=a^2+c^2