3.Trigonometrical Ratios, Functions and Identities
normal

The value of $\frac{{3 + \cot {{76}^o}\cot {{16}^o}}}{{\cot {{76}^o} + \cot {{16}^o}}}$ 

A

$cot\,\, 46^o$

B

$tan\,\, 44^o$

C

$tan\,\, 2^o$

D

$cot\,\, 44^o$

Solution

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

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

$=\frac{2 \sin 76^{\circ} \sin 16^{\circ}+\cos \left(76^{\circ}-16^{\circ}\right)}{\sin \left(76^{\circ}+16^{\circ}\right)}$

$=\frac{2 \sin 76^{\circ} \sin 16^{\circ}+\frac{1}{2}}{\sin \left(92^{\circ}\right)}$

$ = \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)}}$

$=\frac{2 \sin ^{2} 46^{\circ}}{2 \sin 46^{\circ} \cos 46^{\circ}}=\tan \left(46^{\circ}\right)$

$=\cot \left(44^{\circ}\right)$

Standard 11
Mathematics

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