MCQ
$\frac{{3 + \cot \,7\,{6^ \circ }\,\cot \,{{16}^ \circ }}}{{\cot \,{{76}^ \circ } + \cot \,{{16}^ \circ }}}$ =
- A$cot \,44^º$
- B$tan \, 44^º$
- C$tan \, 2^º$
- D$cot \, 46^º$
$=$ $\frac{{2\,\sin {{76}^0}\,\sin {{16}^0}\, + \,[\sin {{76}^0}\,\sin {{16}^0}\, + \,\cos {{76}^0}\,\cos {{16}^0}]}}{{\sin {{92}^0}}}$
$=$ $\frac{{\cos 60^\circ - \cos 92^\circ + \cos 60^\circ }}{{\sin 92^\circ }}$
$=$$\frac{{1 - \cos {{92}^0}}}{{\sin {{92}^0}}}$ $=$$\frac{{2\,{{\sin }^2}{{46}^0}}}{{2\,\sin {{46}^0}\,\cos {{46}^0}}}\,$ $= tan 46^o = cot44^o$
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