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If $\tan A=\dfrac{1-\cos B}{\sin B}$, prove that $\tan 2A=\tan B$.

My effort:

Here

$$\tan A=\frac{1-\cos B}{\sin B}$$ Now

$$\begin{align}\text{L.H.S.} &=\tan 2A \\[4pt] &=\frac{2\tan A}{1-\tan ^2A} \\[6pt] &=\frac{(2-2\cos B)\over\sin B}{1-\frac{(1-\cos B)^2}{\sin^2 B}} \end{align}$$

On simplification from here, I could not get the required R.H.S.

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  • $\begingroup$ Multiply top and bottom by $\sin^2B$, expand the denominator in which you replace $\sin^2B$ by $1-\cos^2B$. Factor and simplify. $\endgroup$ – Claude Leibovici Feb 7 '16 at 12:34
  • $\begingroup$ $1-(1-\cos B)^2/\sin^2B=(\sin^2B-1-\cos^2B+2\cos B)/\sin^2B$ and then remember that $\sin^2B-1=-\cos^2B$. $\endgroup$ – Aretino Feb 7 '16 at 12:36
  • $\begingroup$ All you have to do is write tan A = tan(B/2) and then say it implies, tan 2A = tan B $\endgroup$ – user230452 Feb 7 '16 at 13:01
  • $\begingroup$ It seem that you got stuck on the denominator. Proceed with it as follows. $$1-\frac{(1-\cos B)^2}{\sin B} = \frac{\sin^2B}{\sin^2B}-\frac{(1-\cos B)^2}{\sin^2 B}= \frac{\sin^2 B-(1-\cos B)^2}{\sin^2B}$$ Can you proceed from here? Hint: $\sin^2 B + \cos^2 B=1$ $\endgroup$ – John Joy Feb 7 '16 at 17:03
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You can proceed in this way:

$$\tan A=\frac{1-\cos B}{\sin B}=\frac{2\sin^2 \frac{B}{2}}{2\sin\frac{B}{2}\cos\frac{B}{2}}=\frac{\sin\frac{B}{2}}{\cos\frac{B}{2}}=\tan \frac{B}{2}$$

And hence comparing, we can write that $A=n\pi + \frac{B}{2}$ where $n$ is any integer.

So we can say that $2A=2n\pi + B \Rightarrow \tan 2A = \tan(2n\pi + B) = \tan B$

Hence proved.

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  • $\begingroup$ could you please explain me about the part from"And hence comparing......... I could not understand it... $\endgroup$ – Iaamuser user Feb 7 '16 at 12:43
  • $\begingroup$ @Iaamuseruser Its a kind of empirical formula. If $\tan x= \tan y$ then $x=n\pi +y$. You can check this by taking different values of $x,y$. And after that, I just multiplied both sides of the equation by $2$ and then took the $\tan$ on both sides. $\endgroup$ – SchrodingersCat Feb 7 '16 at 12:49
  • $\begingroup$ @ SchrodingersCat, you have written, "And hence comparing" but comparing with what? $\endgroup$ – Iaamuser user Feb 7 '16 at 12:52
  • $\begingroup$ @SchrodingersCat, $B=0$ needs special handling! Otherwise its my own way:) $\endgroup$ – lab bhattacharjee Feb 7 '16 at 12:56
  • $\begingroup$ @Iaamuseruser By that "And hence comparing", I just compared the R.H.S. with the L.H.S. Nothing else.... No need to stress on that in particular.... :-) $\endgroup$ – SchrodingersCat Feb 7 '16 at 13:02
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Note that the given expression implies:

$\tan(A)=\frac{1-\cos(B)}{\sin(B)} \cdot \frac{1+\cos(B)}{1+\cos(B)}=\frac{\sin(B)}{1+\cos(B)}$

Therefore

$\tan(2A)=\frac{2\tan(A)}{1-\tan^2(A)}=\frac{2\frac{\sin(B)}{1+\cos(B)}}{1-\left(\frac{1-\cos(B)}{\sin(B)}\cdot \frac{\sin(B)}{1+\cos(B)}\right)}=\frac{2\sin(B)}{2\cos(B)}=\tan(B)$

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You are on the right road.... Continuing calculation, \begin{align} \frac{\frac{2-2\cos B}{\sin B}}{1-\frac{(1-\cos B)^2}{\sin^2 B}}&=\frac{\frac{2-2\cos B}{\sin B}}{\frac{\sin^2 B -1+2\cos B-\cos^2 B }{\sin^2 B}}\\ &=\frac{\frac{2-2\cos B}{\sin B}}{\frac{\sin^2 B -\sin^2 B -\cos^2 B+2\cos B-\cos^2 B }{\sin^2 B}}\\ &=\frac{\frac{2-2\cos B}{\sin B}}{\frac{2\cos B(1-\cos B)}{\sin^2 B}}\\ &=\frac{2\sin^2 B(1-\cos B)}{2\cos B\sin B(1-\cos B)}\\ &=\frac{\sin B}{\cos B}\\ &=\tan B \end{align}

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Continuing from your last step, $$ LHS=\frac{\frac{2-2\cos B}{\sin B}}{1-\frac{(1-\cos B)^2}{\sin^2B}}\\ =\frac{\frac{2-2\cos B}{\sin B}}{\frac{\sin^2B-(1-\cos B)^2}{\sin^2B}} \\ =\frac{2-2\cos B}{\frac{\sin^2B-1-\cos^2B+2\cos B}{\sin B}} $$ and then use that $1=\sin^2+\cos^2B$ to get $$ LHS= \frac{2(1-\cos B)(\sin B)}{\sin^2B-\sin^2B-\cos^2B-\cos^2B+2\cos B}\\ =\frac{2(1-\cos B)(\sin B)}{2\cos B(1-\cos B)}\\ =\frac{\sin B}{\cos B}=\tan B=RHS $$

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