cot 5^o - tan 5^o - 2 tan 10^o - 4 tan 20^o - | vedclass

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(A) We use the identity $\cot 	heta - 	an 	heta = 2 \cot 2	heta$. Step $1$: Simplify the first two terms: $\cot 5^o - 	an 5^o = 2 \cot 10^o$. Ste

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(A) We use the identity $\cot 	heta - 	an 	heta = 2 \cot 2	heta$. Step $1$: Simplify the first two terms: $\cot 5^o - 	an 5^o = 2 \cot 10^o$. Step $2$: Substitute this into the expression: $2 \cot 10^o - 2 	an 10^o - 4 	an 20^o - 8 \cot 40^o$. Step $3$: Factor out $2$: $2(\cot 10^o - 	an 10^o) - 4 	an 20^o - 8 \cot 40^o = 2(2 \cot 20^o) - 4 	an 20^o - 8 \cot 40^o = 4 \cot 20^o - 4 	an 20^o - 8 \cot 40^o$. Step $4$: Factor out $4$: $4(\cot 20^o - 	an 20^o) - 8 \cot 40^o = 4(2 \cot 40^o) - 8 \cot 40^o = 8 \cot 40^o - 8 \cot 40^o = 0$.

$cot 5^o - tan 5^o - 2 tan 10^o - 4 tan 20^o - 8 cot 40^o$ is equal to

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