The value of cot 12^{circ} cot 102^{circ} + c | vedclass

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Question

(B) Let the expression be $E = \cot 12^{\circ} \cot 102^{\circ} + \cot 102^{\circ} \cot 66^{\circ} + \cot 66^{\circ} \cot 12^{\circ}$.We know that $\c

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Let the expression be $E = \cot 12^{\circ} \cot 102^{\circ} + \cot 102^{\circ} \cot 66^{\circ} + \cot 66^{\circ} \cot 12^{\circ}$.We know that $\cot 102^{\circ} = \cot(90^{\circ} + 12^{\circ}) = -	an 12^{\circ}$.Substituting this into the expression:$E = \cot 12^{\circ} (-	an 12^{\circ}) + (-	an 12^{\circ}) \cot 66^{\circ} + \cot 66^{\circ} \cot 12^{\circ}$$E = -1 - 	an 12^{\circ} \cot 66^{\circ} + \cot 66^{\circ} \cot 12^{\circ}$$E = -1 + \cot 66^{\circ} (\cot 12^{\circ} - 	an 12^{\circ})$Using the identity $\cot 	heta - 	an 	heta = 2 \cot 2	heta$:$E = -1 + \cot 66^{\circ} (2 \cot 24^{\circ})$Since $\cot 24^{\circ} = \cot(90^{\circ} - 66^{\circ}) = 	an 66^{\circ}$:$E = -1 + 2 \cot 66^{\circ} 	an 66^{\circ}$$E = -1 + 2(1) = 1$.

The value of $\cot 12^{\circ} \cot 102^{\circ} + \cot 102^{\circ} \cot 66^{\circ} + \cot 66^{\circ} \cot 12^{\circ}$ is

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