tan 20^circ cot 10^circ cot 50^circ is equal | vedclass

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(B) We know that $\cot 	heta = 	an(90^\circ - 	heta)$.Given expression: $	an 20^\circ \cot 10^\circ \cot 50^\circ = 	an 20^\circ 	an(90^\circ -

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(B) We know that $\cot 	heta = 	an(90^\circ - 	heta)$.Given expression: $	an 20^\circ \cot 10^\circ \cot 50^\circ = 	an 20^\circ 	an(90^\circ - 10^\circ) 	an(90^\circ - 50^\circ)$.$= 	an 20^\circ 	an 80^\circ 	an 40^\circ$.Rearranging the terms: $	an 20^\circ 	an 40^\circ 	an 80^\circ$.Using the identity $	an 	heta 	an(60^\circ - 	heta) 	an(60^\circ + 	heta) = 	an 3	heta$ where $	heta = 20^\circ$:$= 	an(3 	imes 20^\circ) = 	an 60^\circ = \sqrt{3}$.

$\tan 20^\circ \cot 10^\circ \cot 50^\circ$ is equal to

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