MCQ
The value of $cot\, x + cot\, (60^o + x) + cot\, (120^o + x)$ is equal to :
- A$cot\, 3x$
- B$tan\, 3x$
- C$3\, tan \,3x$
- ✓$\frac{{3\,\, - \,\,9\,{{\tan }^2}\,x}}{{3\,\tan \,x\,\, - \,\,{{\tan }^3}\,x}}$
$cotx +\frac{{\cos (60 + x)}}{{\sin (60 + x)}}\,\, + \,\,\frac{{\cos (x - 60)}}{{\sin (x - 60)}}$
$=$ $\frac{{\cos x}}{{\sin x}}\,\, + \,\,\frac{{\sin (2x)}}{{\sin (x + 60)\,\sin (x - 60)}}$
$=$$\frac{{\cos x}}{{\sin x}}\,\, + \,\,\frac{{8\,\sin x\,\,\cos x}}{{4{{\sin }^2}x\, - \,3}}$
$=$ $\frac{{4\,{{\sin }^2}x\,\,\cos x\,\, - \,3\cos x\,\, + \,8\,{{\sin }^2}x\,\cos x}}{{4{{\sin }^3}x\, - \,3\,\sin x}}$
$=$ $\frac{{3[3\cos x - 4{{\cos }^3}x]}}{{{{\sin }^3}x}}\,\,$
$= 3\, cot3x$
$\Rightarrow$ $\frac{{3[1 - 3{{\tan }^2}x]}}{{3\tan x\, - \,{{\tan }^3}x}}\,\,$
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