tan 20^circ tan 40^circ tan 60^circ tan 80^ci | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We use the identity $	an 	heta 	an(60^\circ - 	heta) 	an(60^\circ + 	heta) = 	an 3	heta$.Given expression is $E = 	an 20^\circ 	an 40^\c

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) We use the identity $	an 	heta 	an(60^\circ - 	heta) 	an(60^\circ + 	heta) = 	an 3	heta$.Given expression is $E = 	an 20^\circ 	an 40^\circ 	an 60^\circ 	an 80^\circ$.Rearranging the terms,we have $E = 	an 60^\circ (	an 20^\circ 	an 40^\circ 	an 80^\circ)$.Note that $	an 80^\circ = 	an(60^\circ + 20^\circ)$ and $	an 40^\circ = 	an(60^\circ - 20^\circ)$.Using the identity for $	heta = 20^\circ$:$	an 20^\circ 	an(60^\circ - 20^\circ) 	an(60^\circ + 20^\circ) = 	an(3 	imes 20^\circ) = 	an 60^\circ$.Substituting this back into the expression for $E$:$E = 	an 60^\circ 	imes 	an 60^\circ = \sqrt{3} 	imes \sqrt{3} = 3$.

$\tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ = $

Similar Questions

$\tan \left( \frac{\pi }{4} + \theta \right) - \tan \left( \frac{\pi }{4} - \theta \right) = $

If $\sin x + \cos x = \frac{1}{5},$ then $\tan 2x$ is

If $A$ is not an integral multiple of $\frac{\pi}{2}$,then $\operatorname{cosec} 2A + \cot 2A$ is equal to

$\tan \left(\frac{3 \pi}{16}\right)+\cot \left(\frac{3 \pi}{16}\right)=$

$\tan A + 2 \tan 2A + 4 \tan 4A + 8 \cot 8A = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module