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

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(D) Let the expression be $E = \frac{3 + \cot 76^{\circ} \cot 16^{\circ}}{\cot 76^{\circ} + \cot 16^{\circ}}$.Converting to sine and cosine,we get $E

Vedclass · Accepted Answer

$\cot 44^{\circ}$

Vedclass · Answer

(D) Let the expression be $E = \frac{3 + \cot 76^{\circ} \cot 16^{\circ}}{\cot 76^{\circ} + \cot 16^{\circ}}$.Converting to sine and cosine,we get $E = \frac{3 + \frac{\cos 76^{\circ} \cos 16^{\circ}}{\sin 76^{\circ} \sin 16^{\circ}}}{\frac{\cos 76^{\circ}}{\sin 76^{\circ}} + \frac{\cos 16^{\circ}}{\sin 16^{\circ}}} = \frac{3 \sin 76^{\circ} \sin 16^{\circ} + \cos 76^{\circ} \cos 16^{\circ}}{\cos 76^{\circ} \sin 16^{\circ} + \sin 76^{\circ} \cos 16^{\circ}}$.Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,the denominator becomes $\sin(76^{\circ} + 16^{\circ}) = \sin 92^{\circ}$.For the numerator,we use $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$:$E = \frac{2 \sin 76^{\circ} \sin 16^{\circ} + \sin 76^{\circ} \sin 16^{\circ} + \cos 76^{\circ} \cos 16^{\circ}}{\sin 92^{\circ}} = \frac{(\cos 60^{\circ} - \cos 92^{\circ}) + \cos(76^{\circ} - 16^{\circ})}{\sin 92^{\circ}}$.Since $\cos 60^{\circ} = \frac{1}{2}$ and $\cos 60^{\circ} = \cos 60^{\circ}$,we simplify the numerator to $\frac{1}{2} - \cos 92^{\circ} + \cos 60^{\circ} = 1 - \cos 92^{\circ}$.Thus,$E = \frac{1 - \cos 92^{\circ}}{\sin 92^{\circ}} = \frac{2 \sin^2 46^{\circ}}{2 \sin 46^{\circ} \cos 46^{\circ}} = 	an 46^{\circ}$.Since $	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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