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

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Question

(A) Given expression: $E = \frac{3 + \cot 76^\circ \cot 16^\circ}{\cot 76^\circ + \cot 16^\circ}$.Converting to sine and cosine: $E = \frac{3 + \frac{

Vedclass · Accepted Answer

$\cot 44^\circ$

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(A) Given expression: $E = \frac{3 + \cot 76^\circ \cot 16^\circ}{\cot 76^\circ + \cot 16^\circ}$.Converting to sine and cosine: $E = \frac{3 + \frac{\cos 76^\circ \cos 16^\circ}{\sin 76^\circ \sin 16^\circ}}{\frac{\cos 76^\circ}{\sin 76^\circ} + \frac{\cos 16^\circ}{\sin 16^\circ}} = \frac{3 \sin 76^\circ \sin 16^\circ + \cos 76^\circ \cos 16^\circ}{\cos 76^\circ \sin 16^\circ + \sin 76^\circ \cos 16^\circ}$.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,write $3 = 2 + 1$: $2 \sin 76^\circ \sin 16^\circ + (\sin 76^\circ \sin 16^\circ + \cos 76^\circ \cos 16^\circ)$.Using $\cos(A-B) = \cos A \cos B + \sin A \sin B$,the term in brackets is $\cos(76^\circ - 16^\circ) = \cos 60^\circ = \frac{1}{2}$.Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$,the first term is $\cos 60^\circ - \cos 92^\circ = \frac{1}{2} - \cos 92^\circ$.Numerator $= \frac{1}{2} - \cos 92^\circ + \frac{1}{2} = 1 - \cos 92^\circ$.Thus,$E = \frac{1 - \cos 92^\circ}{\sin 92^\circ} = \frac{2 \sin^2 46^\circ}{2 \sin 46^\circ \cos 46^\circ} = 	an 46^\circ = \cot(90^\circ - 46^\circ) = \cot 44^\circ$.

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

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