MCQ
$\cot x - \tan x = $
- A$\cot \,2x$
- B$2{\cot ^2}x$
- ✓$2\,\,\cot \,2x$
- D${\cot ^2}\,2x$
$ = \frac{{2\,\cos \,2x}}{{\sin \,2x}} = 2\,\,\cot \,\,2x.$
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