The value of cot, x + cot, (60^o + x) + | vedclass

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Question

Note that the given expression simplifies to $3 \, cot3x$

$cotx +\frac{{\cos (60 + x)}}{{\sin (60 + x)}}\,\, + \,\,\frac{{\cos (x - 60)}}{{\sin (x

Vedclass · Accepted Answer

$\frac{{3\,\, - \,\,9\,{{	an }^2}\,x}}{{3\,	an \,x\,\, - \,\,{{	an }^3}\,x}}$

Vedclass · Answer

Note that the given expression simplifies to $3 \, cot3x$

$cotx +\frac{{\cos (60 + x)}}{{\sin (60 + x)}}\,\, + \,\,\frac{{\cos (x - 60)}}{{\sin (x - 60)}}$

$=$ $\frac{{\cos x}}{{\sin x}}\,\, + \,\,\frac{{\sin (2x)}}{{\sin (x + 60)\,\sin (x - 60)}}$

$=$$\frac{{\cos x}}{{\sin x}}\,\, + \,\,\frac{{8\,\sin x\,\,\cos x}}{{4{{\sin }^2}x\, - \,3}}$ 

$=$ $\frac{{4\,{{\sin }^2}x\,\,\cos x\,\, - \,3\cos x\,\, + \,8\,{{\sin }^2}x\,\cos x}}{{4{{\sin }^3}x\, - \,3\,\sin x}}$

$=$ $\frac{{3[3\cos x - 4{{\cos }^3}x]}}{{{{\sin }^3}x}}\,\,$

$= 3\, cot3x$

$\Rightarrow$ $\frac{{3[1 - 3{{	an }^2}x]}}{{3	an x\, - \,{{	an }^3}x}}\,\,$

The value of $cot\, x + cot\, (60^o + x) + cot\, (120^o + x)$ is equal to :

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