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Proving (if $x + y +z = 90$) -

$$\cos x (\cos (y - z) - \cos (y +z)) + \cos y (\cos (x - z) - \cos (x + z)) + \cos z (\cos (x -y)- cos (x + y)) = 2 \cos x \cos y \cos z$$

I've no idea about the solution of the Equation. I can't take common and the difference inside the cosine ratios are difficult to be removed.

Question - Prove that -

$$\cot\frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac {C}{2}$$ When $A + B + C = 180^o$

I tried to solve it like this here -

Let $\frac{A}{2} = x$, $\frac{B}{2} = y$, and $\frac{C}{2} = z$

Let $\alpha$ = $\cot x + \cot y + \cot z$ $$\Rightarrow \alpha = \frac{\sin y \sin z \cos x + \sin x \cos y \sin z + \sin x \sin y \cos z}{ \sin x \sin y \sin z}$$ $$\Rightarrow \alpha = \frac{\cos x (\cos (y - z) - \cos (y +z)) + \cos y (\cos (x - z) - \cos (x + z)) + \cos z (\cos (x -y)- cos (x + y))}{2 \sin x \sin y \sin z}$$

Now, If I'd be able to prove the numerator as $2 \cos x \cos y \cos z$, then I'd be able to make the whole equation = $\cot \frac{A}{2} \cot \frac{B}{2} \cot \frac {C}{2}$

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  • $\begingroup$ Do you want to prove that $\cot{a/2}$+$\cot{b/2}$+$\cot{c/2}$ = $\cot{a/2}$$\cot{b/2}$$\cot{c/2}$ $\endgroup$ – Haran May 31 '18 at 15:14
  • $\begingroup$ @Haran Why are you repeating the question? $\endgroup$ – TheSimpliFire May 31 '18 at 15:15
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We have: $$x+y = \frac{π}{2} - z$$

Then, $$\tan{(x+y)} = \tan{({{{\frac{π}{2}}-z}})}$$

$$\frac{\cot{x}+\cot{y}}{\cot{x}\cot{y}-1} = \cot{z}$$

Hence, when $x+y+z=\frac{π}{2}$, we have:

$$\cot{x}+\cot{y}+\cot{z}=\cot{x}\cot{y}\cot{z}$$

Now, replace $x=\frac{A}{2}, y=\frac{B}{2}, z=\frac{C}{2}$, $$\cot{\frac{A}{2}}+\cot{\frac{B}{2}}+\cot{\frac{C}{2}}=\cot{\frac{A}{2}}\cot{\frac{B}{2}}\cot{\frac{C}{2}}$$

Hence, proved.

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  • $\begingroup$ (+1) Brilliant! $\endgroup$ – TheSimpliFire May 31 '18 at 15:36
  • $\begingroup$ @haran can you try to do this in my way? $\endgroup$ – Abhas Kumar Sinha May 31 '18 at 15:41
  • $\begingroup$ tan (x + y) changes to cot in next step? $\endgroup$ – Abhas Kumar Sinha May 31 '18 at 15:43
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Hint: use the formulas $$\cot\left(\frac{\alpha}{2}\right)=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$$ etc Then Show that $$\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}+\sqrt{\frac{s(s-b)}{(s-a)(s-c)}}+\sqrt{\frac{s(s-c)}{(s-a)(s-b)}}=\sqrt{\frac{s(s-a)s(s-b)s(s-c)}{(s-b)(s-c)(s-a)(s-c)(s-a)(s-b)}}$$ Write the left-hand side as: $$\sqrt{\frac{s^2(s-a)^2}{s(s-a)(s-b)(s-c)}}+\sqrt{\frac{s^2(s-b)^2}{s(s-a)(s-b)(s-c)}}+\sqrt{\frac{s^2(s-c)^2}{s(s-a)(s-b)(s-c)}}$$ and this is $$\frac{1}{A}(s(s-a)+s(s-b)+s(s-c))$$

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  • $\begingroup$ I've not seen any formula like that which you are using, isn't there a good alternative? $\endgroup$ – Abhas Kumar Sinha May 31 '18 at 15:16
  • $\begingroup$ I've not read $$\cot \frac{\alpha}{2} = \sqrt({\frac{s(s -a)}{(s-b)(s-c)}})$$ also, there is a $\alpha$ in LHS but not in RHS, so, equation seems incorrect $\endgroup$ – Abhas Kumar Sinha May 31 '18 at 15:18
  • $\begingroup$ This is absolutely correct, it called the half angle formulas! $\endgroup$ – Dr. Sonnhard Graubner May 31 '18 at 15:19
  • $\begingroup$ I can't use it in the exams because I've not read that $\endgroup$ – Abhas Kumar Sinha May 31 '18 at 15:20
  • $\begingroup$ How it's correct? If you can say that it's correct, then I can also say that $$\sin 1^o = \sqrt{a + b}$$ $\endgroup$ – Abhas Kumar Sinha May 31 '18 at 15:21
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In a triangle $A+B+C=\pi$ $$\pi-C=A+B$$ Divide by $2$ on both sides $$\frac\pi2-\frac C2=\frac A2+\frac B2$$ $$\tan\left(\frac\pi2-\frac C2\right)=\tan\left(\frac A2+\frac B2\right)$$ $$\cot\left(\frac C2\right)=\frac{\left(\tan \frac A2+\tan\frac B2\right)}{\left(1-\tan\frac A2\tan\frac B2\right)}$$ $$\cot\left(\frac C2\right)=\frac{\cot\frac B2+\cot\frac A2}{\cot\frac A2\cot\frac B2-1}$$ $$\cot\frac A2\cot\frac B2\cot\frac C2-\cot\frac C2=\cot\frac A2+\cot\frac B2+\cot\frac C2$$

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