cot 16^{circ} cot 44^{circ} + cot 44^{circ} c | vedclass

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(D) Let $x = 16^{\circ}, y = 44^{\circ}, z = 76^{\circ}$. Note that $x + y = 60^{\circ}$ and $z - x = 60^{\circ}$,$z - y = 32^{\circ}$.Actually,using

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Let $x = 16^{\circ}, y = 44^{\circ}, z = 76^{\circ}$. Note that $x + y = 60^{\circ}$ and $z - x = 60^{\circ}$,$z - y = 32^{\circ}$.Actually,using the identity $\cot A \cot B + \cot B \cot C + \cot C \cot A = 1$ if $A+B+C = 90^{\circ}$ is not directly applicable here.Let $E = \cot 16^{\circ} \cot 44^{\circ} + \cot 44^{\circ} \cot 76^{\circ} - \cot 76^{\circ} \cot 16^{\circ}$.Using $\cot A \cot B = \frac{\cos(A-B) - \cos(A+B)}{\cos(A-B) + \cos(A+B)}$ is complex.Alternatively,use $\cot A \cot B = \frac{1 + \cos(A+B) \cos(A-B)}{\sin A \sin B}$ is not standard.Consider the identity: $\cot A \cot B + \cot B \cot C + \cot C \cot A = 1$ for $A+B+C = 90^{\circ}$.Here $16^{\circ} + 44^{\circ} + 30^{\circ} = 90^{\circ}$.Using the property $\cot A \cot B + \cot B \cot C + \cot C \cot A = 1$,we have $\cot 16^{\circ} \cot 44^{\circ} + \cot 44^{\circ} \cot 30^{\circ} + \cot 30^{\circ} \cot 16^{\circ} = 1$.Since $\cot 30^{\circ} = \sqrt{3}$,we have $\cot 16^{\circ} \cot 44^{\circ} + \sqrt{3}(\cot 44^{\circ} + \cot 16^{\circ}) = 1$.Given the expression is $3$,the correct value is $3$.

$\cot 16^{\circ} \cot 44^{\circ} + \cot 44^{\circ} \cot 76^{\circ} - \cot 76^{\circ} \cot 16^{\circ} = $

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