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# If tanA+tanB=x & cotA+cotB=y, then prove that-

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If tanA+tanB=x & cotA+cotB=y, then prove that-

cot(A+B)=1/x-1/y

Mar 29, 2016

#1
+26172
+15

If tanA+tanB=x & cotA+cotB=y, then prove that-

cot(A+B)=1/x-1/y

$$\begin{array}{rcll} \tan(A) + \tan(B) &=& x \\ \cot(A) + \cot(B) &=& y \end{array}$$

formulary

$$\begin{array}{rcll} \tan(A+B) &=& \dfrac{\tan(A) + \tan(B)}{1-\tan(A) \cdot \tan(B)} \\\\ \cot(A+B) &=& \dfrac{ \cot(A) \cdot \cot(B) - 1 } {\cot(A) + \cot(B)} \end{array}$$

1.

$$\begin{array}{rcll} \tan(A+B)=\dfrac{1}{\cot(A+B)} &=& \dfrac{x} {1-\tan(A) \cdot \tan(B)} \\\\ 1-\tan(A) \cdot \tan(B) &=& x\cdot \cot(A+B) \\\\ \tan(A) \cdot \tan(B)-1 &=& -x\cdot \cot(A+B) \\\\ \tan(A) \cdot \tan(B) &=& 1 -x\cdot \cot(A+B) \\\\ \dfrac{1}{\cot(A)} \cdot \dfrac{1}{\cot(B)} &=& 1 -x\cdot \cot(A+B) \\\\ \mathbf{ \cot(A) \cdot \cot(B) } &\mathbf{=}& \mathbf{ \dfrac{1}{ 1 -x\cdot \cot(A+B) } } \end{array}$$

2.

$$\begin{array}{rcll} \cot(A+B) &=& \dfrac{ \cot(A) \cdot \cot(B) - 1 } {y} \\\\ y\cdot \cot(A+B) &=& \cot(A) \cdot \cot(B) - 1 \\\\ 1+y\cdot \cot(A+B) &=& \cot(A) \cdot \cot(B) \\ && \cot(A) \cdot \cot(B) =\dfrac{1}{ 1 -x\cdot \cot(A+B) } \\ 1+y\cdot \cot(A+B) &=& \dfrac{1}{ 1 -x\cdot \cot(A+B) } \\ [ 1+y\cdot \cot(A+B)]\cdot[1 -x\cdot \cot(A+B)] &=& 1 \\ 1 -x\cdot \cot(A+B) + y\cdot \cot(A+B) - yx\cdot \cot^2(A+B) &=& 1 \\ -x\cdot \cot(A+B) + y\cdot \cot(A+B) - yx\cdot \cot^2(A+B) &=& 0 \quad | \quad : \cot(A+B)\\ -x + y - yx\cdot \cot(A+B) &=& 0 \\ yx\cdot \cot(A+B) &=& y -x \quad | \quad : xy\\ \cot(A+B) &=& \dfrac{y -x}{xy} \\ \cot(A+B) &=& \dfrac{y}{xy}-\dfrac{x}{xy} \\ \mathbf{ \cot(A+B)} &\mathbf{ =}&\mathbf{ \dfrac{1}{x}-\dfrac{1}{y} } \end{array}$$

Mar 29, 2016

#1
+26172
+15

If tanA+tanB=x & cotA+cotB=y, then prove that-

cot(A+B)=1/x-1/y

$$\begin{array}{rcll} \tan(A) + \tan(B) &=& x \\ \cot(A) + \cot(B) &=& y \end{array}$$

formulary

$$\begin{array}{rcll} \tan(A+B) &=& \dfrac{\tan(A) + \tan(B)}{1-\tan(A) \cdot \tan(B)} \\\\ \cot(A+B) &=& \dfrac{ \cot(A) \cdot \cot(B) - 1 } {\cot(A) + \cot(B)} \end{array}$$

1.

$$\begin{array}{rcll} \tan(A+B)=\dfrac{1}{\cot(A+B)} &=& \dfrac{x} {1-\tan(A) \cdot \tan(B)} \\\\ 1-\tan(A) \cdot \tan(B) &=& x\cdot \cot(A+B) \\\\ \tan(A) \cdot \tan(B)-1 &=& -x\cdot \cot(A+B) \\\\ \tan(A) \cdot \tan(B) &=& 1 -x\cdot \cot(A+B) \\\\ \dfrac{1}{\cot(A)} \cdot \dfrac{1}{\cot(B)} &=& 1 -x\cdot \cot(A+B) \\\\ \mathbf{ \cot(A) \cdot \cot(B) } &\mathbf{=}& \mathbf{ \dfrac{1}{ 1 -x\cdot \cot(A+B) } } \end{array}$$

2.

$$\begin{array}{rcll} \cot(A+B) &=& \dfrac{ \cot(A) \cdot \cot(B) - 1 } {y} \\\\ y\cdot \cot(A+B) &=& \cot(A) \cdot \cot(B) - 1 \\\\ 1+y\cdot \cot(A+B) &=& \cot(A) \cdot \cot(B) \\ && \cot(A) \cdot \cot(B) =\dfrac{1}{ 1 -x\cdot \cot(A+B) } \\ 1+y\cdot \cot(A+B) &=& \dfrac{1}{ 1 -x\cdot \cot(A+B) } \\ [ 1+y\cdot \cot(A+B)]\cdot[1 -x\cdot \cot(A+B)] &=& 1 \\ 1 -x\cdot \cot(A+B) + y\cdot \cot(A+B) - yx\cdot \cot^2(A+B) &=& 1 \\ -x\cdot \cot(A+B) + y\cdot \cot(A+B) - yx\cdot \cot^2(A+B) &=& 0 \quad | \quad : \cot(A+B)\\ -x + y - yx\cdot \cot(A+B) &=& 0 \\ yx\cdot \cot(A+B) &=& y -x \quad | \quad : xy\\ \cot(A+B) &=& \dfrac{y -x}{xy} \\ \cot(A+B) &=& \dfrac{y}{xy}-\dfrac{x}{xy} \\ \mathbf{ \cot(A+B)} &\mathbf{ =}&\mathbf{ \dfrac{1}{x}-\dfrac{1}{y} } \end{array}$$

heureka Mar 29, 2016
#2
+5

Here's an alternative method.

$$\displaystyle \cot(A+B)=\frac{1}{\tan(A+B)}=\frac{1-\tan(A)\tan(B)}{\tan(A)+\tan(B)} \\ \displaystyle =\frac{(\cot(A)+\cot(B))(1-\tan(A)\tan(B))}{(\cot(A)+\cot(B))(\tan(A)+\tan(B))} \\ \displaystyle =\frac{\cot(A)+\cot(B)-\tan(A)-\tan(B))}{(\cot(A)+\cot(B))(\tan(A)+\tan(B)} \\ \displaystyle=\frac{1}{\tan(A)+\tan(B)}-\frac{1}{\cot(A)+\cot(B)}=\frac{1}{x}-\frac{1}{y}.$$

Mar 29, 2016
#3
+121008
+5

Here's the proof in the opposite direction :

1/x  - 1/y    =

1/ [ tanA + tanB]  - 1 /[ cotA+cotB]

[(cotA + cotB) - (tanA + tanB)] / [(tanA + tanB) ( cotA + cotB)]

[(cotAcotB - 1)/(cot(A + B) - [tan(A + B)(1 - tanAtanB) ] / [(tanA + tanB) ( cotA + cotB)]

[ (cotAcotB -1) - cot(A +B)tan(A+B)( 1 - tanAtanB)]/ [ (cot(A+B)(tanA + tanB)(cotA + cotB)]

[(cotAcotB -1 ) (1 - tanAtanB)] / [(tanA + tanB)(cotAcotB -1)]

(1 - tanAtanB) / (tanA + tanB)  =

1 / [( tanA + tanB) / (1 - tanAtanB)]  =

1 / tan(A + B)   =

cot(A + B)

Mar 29, 2016