cot theta - tan theta - 2 tan 2 theta - 4 tan | vedclass

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Question

(A) We know that $\cot 	heta - 	an 	heta = \frac{\cos 	heta}{\sin 	heta} - \frac{\sin 	heta}{\cos 	heta} = \frac{\cos^2 	heta - \sin^2 	heta}

Vedclass · Accepted Answer

$8 \cot 8 	heta$

Vedclass · Answer

(A) We know that $\cot 	heta - 	an 	heta = \frac{\cos 	heta}{\sin 	heta} - \frac{\sin 	heta}{\cos 	heta} = \frac{\cos^2 	heta - \sin^2 	heta}{\sin 	heta \cos 	heta} = \frac{\cos 2 	heta}{\frac{1}{2} \sin 2 	heta} = 2 \cot 2 	heta$. Substituting this into the expression: $\cot 	heta - 	an 	heta - 2 	an 2 	heta - 4 	an 4 	heta = 2 \cot 2 	heta - 2 	an 2 	heta - 4 	an 4 	heta$. Now,$2(\cot 2 	heta - 	an 2 	heta) = 2(2 \cot 4 	heta) = 4 \cot 4 	heta$. Substituting this back: $4 \cot 4 	heta - 4 	an 4 	heta = 4(\cot 4 	heta - 	an 4 	heta) = 4(2 \cot 8 	heta) = 8 \cot 8 	heta$.

$\cot \theta - \tan \theta - 2 \tan 2 \theta - 4 \tan 4 \theta = $

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