MCQ
$\frac{{3 + \cot {{76}^o}\cot {{16}^o}}}{{\cot {{76}^o} + \cot {{16}^o}}}$ =
- A$cot\,\, 46^o$
- B$tan\,\, 44^o$
- C$tan\,\, 2^o$
- ✓$cot\,\, 44^o$
$=\frac{3 \sin 76^{\circ} \sin 16^{\circ}+\cos 76^{\circ} \cos 16^{\circ}}{\cos 76^{\circ} \sin 16^{\circ}+\sin 76^{\circ} \cos 16^{\circ}}$
$=\frac{2 \sin 76^{\circ} \sin 16^{\circ}+\cos \left(76^{\circ}-16^{\circ}\right)}{\sin \left(76^{\circ}+16^{\circ}\right)}$
$=\frac{2 \sin 76^{\circ} \sin 16^{\circ}+\frac{1}{2}}{\sin \left(92^{\circ}\right)}$
$ = \frac{{\cos {{60}^\circ } - \cos {{92}^\circ } + \frac{1}{2}}}{{\sin \left( {{{92}^\circ }} \right)}} = \frac{{1 - \cos {{92}^\circ }}}{{\sin \left( {{{92}^\circ }} \right)}}$
$=\frac{2 \sin ^{2} 46^{\circ}}{2 \sin 46^{\circ} \cos 46^{\circ}}=\tan \left(46^{\circ}\right)$
$=\cot \left(44^{\circ}\right)$
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