The value of frac{3 + cot 76^{circ} cot 16^{c | vedclass

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(D) Given expression: $E = \frac{3 + \cot 76^{\circ} \cot 16^{\circ}}{\cot 76^{\circ} + \cot 16^{\circ}}$Convert to $\sin$ and $\cos$: $E = \frac{3 +

Vedclass · Accepted Answer

$\cot 44^{\circ}$

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(D) Given expression: $E = \frac{3 + \cot 76^{\circ} \cot 16^{\circ}}{\cot 76^{\circ} + \cot 16^{\circ}}$Convert to $\sin$ and $\cos$: $E = \frac{3 + \frac{\cos 76^{\circ} \cos 16^{\circ}}{\sin 76^{\circ} \sin 16^{\circ}}}{\frac{\cos 76^{\circ}}{\sin 76^{\circ}} + \frac{\cos 16^{\circ}}{\sin 16^{\circ}}} = \frac{3 \sin 76^{\circ} \sin 16^{\circ} + \cos 76^{\circ} \cos 16^{\circ}}{\cos 76^{\circ} \sin 16^{\circ} + \sin 76^{\circ} \cos 16^{\circ}}$Using $\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]$ and $\sin(A+B) = \sin A \cos B + \cos A \sin B$:Numerator: $2 \sin 76^{\circ} \sin 16^{\circ} + (\sin 76^{\circ} \sin 16^{\circ} + \cos 76^{\circ} \cos 16^{\circ}) = 2 \sin 76^{\circ} \sin 16^{\circ} + \cos(76^{\circ} - 16^{\circ}) = 2 \sin 76^{\circ} \sin 16^{\circ} + \cos 60^{\circ} = 2 \sin 76^{\circ} \sin 16^{\circ} + \frac{1}{2}$Denominator: $\sin(76^{\circ} + 16^{\circ}) = \sin 92^{\circ}$Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$:$E = \frac{\cos(76^{\circ}-16^{\circ}) - \cos(76^{\circ}+16^{\circ}) + \frac{1}{2}}{\sin 92^{\circ}} = \frac{\cos 60^{\circ} - \cos 92^{\circ} + \frac{1}{2}}{\sin 92^{\circ}} = \frac{\frac{1}{2} - \cos 92^{\circ} + \frac{1}{2}}{\sin 92^{\circ}} = \frac{1 - \cos 92^{\circ}}{\sin 92^{\circ}}$Using half-angle formulas $1 - \cos 	heta = 2 \sin^2(\frac{	heta}{2})$ and $\sin 	heta = 2 \sin(\frac{	heta}{2}) \cos(\frac{	heta}{2})$:$E = \frac{2 \sin^2 46^{\circ}}{2 \sin 46^{\circ} \cos 46^{\circ}} = 	an 46^{\circ} = \cot(90^{\circ} - 46^{\circ}) = \cot 44^{\circ}$

The value of $\frac{3 + \cot 76^{\circ} \cot 16^{\circ}}{\cot 76^{\circ} + \cot 16^{\circ}}$ is:

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