# Prove that cot (A/2)- 3cot ((3A)/2) = (4sinA)/(1+2cosA)?

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dk_ch Share
Feb 21, 2018

$L H S = \cot \left(\frac{A}{2}\right) - 3 \cot \left(\frac{3 A}{2}\right)$

$= \cos \frac{\frac{A}{2}}{\sin} \left(\frac{A}{2}\right) - \cos \frac{\frac{3 A}{2}}{\sin} \left(\frac{3 A}{2}\right) - 2 \cot \left(\frac{3 A}{2}\right)$

$= \frac{\sin \left(\frac{3 A}{2}\right) \cdot \cos \left(\frac{A}{2}\right) - \cos \left(\frac{3 A}{2}\right) \cdot \sin \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right) \cdot \sin \left(\frac{3 A}{2}\right)} - 2 \cot \left(\frac{3 A}{2}\right)$

=sin(A)/(sin(A/2)*sin((3A)/2)) -2cot((3A)/2)

=(2sin(A/2)cos(A/2))/(sin(A/2)*sin((3A)/2)) -2cot((3A)/2)

=2cos(A/2)/sin((3A)/2)-2*cos((3A)/2)/sin((3 A)/2)

=2[(cos(A/2)-cos((3A)/2))/sin((3 A)/2)]

=2[(2sin(A/2)sin(A))/(3sin( A/2)-4sin^3(A/2))]

=(4sin(A/2)sin(A))/(sin(A/2)(3-4sin^2(A/2))
=(4sin(A))/(3-2(1-cosA))

$= \frac{4 \sin \left(A\right)}{1 + 2 \cos A} = R H S$

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Feb 20, 2018

#### Explanation:

$L H S = \cot \left(\frac{x}{2}\right) - 3 \cot \left(\frac{3 x}{2}\right)$

$= \cos \frac{\frac{x}{2}}{\sin} \left(\frac{x}{2}\right) - 3 \cdot \cos \frac{\frac{3 x}{2}}{\sin} \left(\frac{3 x}{2}\right)$

=(sin((3x)/2)*cos(x/2)-3*cos((3x)/2)*sin(x/2))/(sin(x/2)*sin((3x)/2)

=(2sin((3x)/2)*cos(x/2)-3*2cos((3x)/2)*sin(x/2))/(2sin(x/2)*sin((3x)/2)

=(sin((3x)/2+x/2)+sin((3x)/2-x/2)-3*{sin((3x)/2+x/2)-sin((3x)/2-x/2)})/(cos((3x)/2-x/2)-cos((3x)/2+x/2)

=(sin((4x)/2)+sin((2x)/2)-3*{sin((4x)/2)-sin((2x)/2)})/(cos((2x)/2)-cos((4x)/2)

$= \frac{\sin 2 x + \sin x - 3 \sin 2 x + 3 \sin x}{\cos x - \cos 2 x}$

$= \frac{4 \sin x - 2 \sin 2 x}{\cos x - \left({\cos}^{2} x - {\sin}^{2} x\right)}$

$= \frac{4 \sin x - 4 \sin x \cdot \cos x}{\cos x - {\cos}^{2} x + {\sin}^{2} x}$

$= \frac{4 \sin x \left(1 - \cos x\right)}{\cos x \left(1 - \cos x\right) + \left(1 - \cos x\right) \left(1 + \cos x\right)}$

=(4sinx(1-cosx))/((1-cosx)(cosx+1+cosx)

$= \frac{4 \sin x}{1 + 2 \cos x} = R H S$

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#### Explanation

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Feb 20, 2018

#### Explanation:

We know that,

$\tan 3 \theta = \frac{3 \tan \theta - {\tan}^{3} \theta}{1 - 3 {\tan}^{2} \theta}$.

$\therefore \cot 3 \theta = \frac{1}{\tan 3 \theta} = \frac{1 - 3 {\tan}^{2} \theta}{3 \tan \theta - {\tan}^{3} \theta}$

$\therefore \cot \left(\frac{3 A}{2}\right) = \frac{1 - 3 {\tan}^{2} \left(\frac{A}{2}\right)}{3 \tan \left(\frac{A}{2}\right) - {\tan}^{3} \left(\frac{A}{2}\right)}$.

Letting $\tan \left(\frac{A}{2}\right) = t ,$ we have,

$\cot \left(\frac{A}{2}\right) - 3 \cot \left(\frac{3 A}{2}\right)$,

$= \frac{1}{t} - 3 \left\{\frac{1 - 3 {t}^{2}}{3 t - {t}^{3}}\right\}$,

$\frac{1}{t} - \frac{3 \left(1 - 3 {t}^{2}\right)}{t \left(3 - {t}^{2}\right)}$,

$= \frac{\left(3 - {t}^{2}\right) - 3 \left(1 - 3 {t}^{2}\right)}{t \left(3 - {t}^{2}\right)}$,

$= \frac{8 {t}^{\cancel{2}}}{\cancel{t} \left(3 - {t}^{2}\right)}$,

$= \frac{8 t}{\left(1 + {t}^{2}\right) + 2 \left(1 - {t}^{2}\right)}$

$= \frac{4 \cdot \frac{2 t}{1 + {t}^{2}}}{\frac{1 + {t}^{2}}{1 + {t}^{2}} + 2 \cdot \frac{1 - {t}^{2}}{1 + {t}^{2}}}$.

Note that, $\frac{2 t}{1 + {t}^{2}} = \frac{2 \tan \left(\frac{A}{2}\right)}{1 + {\tan}^{2} \left(\frac{A}{2}\right)} = \sin A , \mathmr{and}$

$\frac{1 - {t}^{2}}{1 + {t}^{2}} = \cos A$.

$\Rightarrow \cot \left(\frac{A}{2}\right) - 3 \cot \left(\frac{3 A}{2}\right) = \frac{4 \sin A}{1 + 2 \cos A} , \text{ as desired!}$

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