frac{{3 + cot {{76}^o}cot {{16}^o}}}{{cot {{ | vedclass

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Question

$\frac{3+\cot 76^{\circ} \cot 16^{\circ}}{\cot 76^{\circ}+\cot 16^{\circ}}=\frac{3+\frac{\cos 76^{\circ} \cos 16^{\circ}}{\sin 76^{\circ} \sin 16^{\ci

Vedclass · Accepted Answer

$cot\,\, 44^o$

Vedclass · Answer

$\frac{3+\cot 76^{\circ} \cot 16^{\circ}}{\cot 76^{\circ}+\cot 16^{\circ}}=\frac{3+\frac{\cos 76^{\circ} \cos 16^{\circ}}{\sin 76^{\circ} \sin 16^{\circ}}}{\frac{\cos 76^{\circ}}{\sin 76^{\circ}}+\frac{\cos 16^{\circ}}{\sin 16^{\circ}}}$

$=\frac{3 \sin 76^{\circ} \sin 16^{\circ}+\cos 76^{\circ} \cos 16^{\circ}}{\cos 76^{\circ} \sin 16^{\circ}+\sin 76^{\circ} \cos 16^{\circ}}$

$=\frac{2 \sin 76^{\circ} \sin 16^{\circ}+\cos \left(76^{\circ}-16^{\circ}ight)}{\sin \left(76^{\circ}+16^{\circ}ight)}$

$=\frac{2 \sin 76^{\circ} \sin 16^{\circ}+\frac{1}{2}}{\sin \left(92^{\circ}ight)}$

$ = \frac{{\cos {{60}^\circ } - \cos {{92}^\circ } + \frac{1}{2}}}{{\sin \left( {{{92}^\circ }} ight)}} = \frac{{1 - \cos {{92}^\circ }}}{{\sin \left( {{{92}^\circ }} ight)}}$

$=\frac{2 \sin ^{2} 46^{\circ}}{2 \sin 46^{\circ} \cos 46^{\circ}}=	an \left(46^{\circ}ight)$

$=\cot \left(44^{\circ}ight)$

$\frac{{3 + \cot {{76}^o}\cot {{16}^o}}}{{\cot {{76}^o} + \cot {{16}^o}}}$ =

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