## A solution to a S L Loney Part I trig problem for IITJEE Advanced Math

Exercise XXVII. Problem 30.

If a, b, c are in AP, prove that $\cos{A}\cot{A/2}$, $\cos{B}\cot{B/2}$, $\cos{C}\cot{C/2}$ are in AP.

Proof:

Given that $b-a=c-b$

TPT: $\cos{B}\cot{B/2}-\cos{A}\cot{A/2}=\cot{C/2}\cos{C}-\cos{B}\cot{B/2}$. —— Equation 1

Let us try to utilize the following formulae:

$\cos{2\theta}=2\cos^{2}{\theta}-1$ which implies the following:

$\cos{B}=2\cos^{2}(B/2)-1$ and $\cos{A}=2\cos^{2}(A/2)-1$

Our strategy will be reduce LHS and RHS of Equation I to a common expression/value.

$LHS=(\frac{2s(s-b)}{ac}-1)(\frac{\sqrt{\frac{(s)(s-b)}{ac}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}})-\cos{A}\cot{(A/2)}$

which is equal to

$(\frac{2s(s-b)}{ac}-1))(\frac{\sqrt{\frac{(s)(s-b)}{ac}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}})-(2\cos^{2}(A/2)-1)\frac{\cos{A/2}}{\sin{A/2}}$

which is equal to

$(\frac{2s(s-b)}{ac}-1))(\frac{\sqrt{\frac{(s)(s-b)}{ac}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}})-(\frac{2s(s-a)}{bc}-1)\frac{\sqrt{\frac{s(s-a)}{bc}}}{\sqrt{\frac{(s-b)(s-c)}{bc}}}$

which is equal to

$(\frac{2s(s-b)}{ac}-1))\sqrt{s(s-b)}{(s-a)(s-c)}-(\frac{2s(s-a)}{bc}-1)\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$

which in turn equals

$\sqrt{\frac{s}{(s-a)(s-b)(s-c)}}((\frac{2s(s-b)}{ac}-1)(s-b)-(\frac{2s(s-a)}{bc}-1)(s-a))$

From the above, consider only the expression, given below. We will see what it simplifies to:

$\frac{2s(s-b)^{2}}{ac}-(s-b)-\frac{2s(s-a)^{2}}{bc}+(s-a)$

$=\frac{2s(s-b)^{2}}{ac}-\frac{2s(s-a)^{2}}{bc}+b-a$

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

$=\frac{2s(s-b)^{2}}{ca}-\frac{2s(s-a)^{2}}{bc}+c-b$ —- Equation II.

Now, consider RHS of Equation I. Let us see if it also boils down to the above expression after simplification.

$RHS=\cot{(C/2)}\cos{C}-\cos{B}\cot{(B/2)}$

$=(2\cos^{2}{(C/2)}-1)\cot{(C/2)}-(2\cos^{2}({B/2})-1)\cot{(B/2)}$

$=(\frac{2s(s-c)}{ab}-1)\frac{\sqrt{\frac{s(s-c)}{ab}}}{\sqrt{\frac{(s-b)(s-a)}{ab}}}-(\frac{2s(s-b)}{ac})\frac{\sqrt{\frac{s(s-b)}{ac}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}}$

$= \sqrt{\frac{s}{(s-a)(s-b)(s-c)}}((\frac{2s(s-c)}{ab}-1)(s-c)-(\frac{2s(s-b)}{ac}-1)(s-b))$

From equation II and above, what we want is given below:

$\frac{2s(s-c)^{2}}{ab}-(s-c)-\frac{2s(s-b)^{2}}{ac}+(s-b)=\frac{2s(s-b)^{2}}{ac}-\frac{2s(s-a)^{2}}{bc}+c-b$

that is, want to prove that $c(s-c)^{2}+a(s-a)^{2}=2b(s-b)^{2}$

but, it is given that $a+c=2b$ and hence, $c=2b-a$, which means $a+c-b=b$ and $b-a=c-b$

that is, want to prove that

$c(a+b-c)^{2}+a(b+c-a)^{2}=2b(a+c-b)^{2}=2b^{3}$

i.e., want: $c(a+b-c)^{2}+a(b+c-a)^{2}=2b^{3}$

i.e., want: $(2b-a)(a+b-2b+a)^{2}+a(b-a+2b-a)^{2}=2b^{3}$

i.e., want: $(2b-a)(2a-b)^{2}+a(3b-2a)^{2}=2b^{3}$

Now, in the above, $LHS=(2b-a)(4a^{2}+b^{2}-4ab)+a(9b^{2}+4a^{2}-12ab)$

$= 8a^{2}b+2b^{3}-8ab^{2}-4a^{3}-ab^{2}+4ab^{2}+9ab^{2}+4a^{3}-12a^{2}b$

$= 2b^{3}$.

Hence, $LHS+RHS$.

QED.

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