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 = $\cot \frac{A}{2} \cot \frac{B}{2} \cot \frac {C}{2}$

  • $\begingroup$ 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

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:


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.

  • $\begingroup$ (+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

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 $$\frac{1}{A}(s(s-a)+s(s-b)+s(s-c))$$

  • $\begingroup$ 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

In a triangle $A+B+C=\pi$ $$\pi-C=A+B$$ Divide by $2$ on both sides $$\frac\pi2-\frac C2=\frac A2+\frac B2$$ $$\tan\left(\frac\pi2-\frac C2\right)=\tan\left(\frac A2+\frac B2\right)$$ $$\cot\left(\frac C2\right)=\frac{\left(\tan \frac A2+\tan\frac B2\right)}{\left(1-\tan\frac A2\tan\frac B2\right)}$$ $$\cot\left(\frac C2\right)=\frac{\cot\frac B2+\cot\frac A2}{\cot\frac A2\cot\frac B2-1}$$ $$\cot\frac A2\cot\frac B2\cot\frac C2-\cot\frac C2=\cot\frac A2+\cot\frac B2+\cot\frac C2$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.