Prove the following identity: `tan θ tan(60^@ +θ)tan(60^@ −θ)=]tan3θ and hence tan20^@ tan40^@ tan60^@ tan80^@ =3.`

Explanation

As we know that


`sin theta sin (60^{circ}-theta) sin (60^{circ}+theta)=frac{1}{4} sin 3 theta`


`cos theta cos (60^{circ}-theta) cos (60^{circ}+theta)=frac{1}{4} cos 3 theta`


`tan theta tan (60^{circ}-theta) tan (60^{circ}+theta)=frac{sin theta}{cos theta} times frac{sin (60^{circ}-theta)}{cos (60^{circ}-theta)} times frac{sin (60^{circ}+theta)}{cos (60^{circ}+theta)}`


`=frac{sin theta sin (60^{circ}-theta) sin (60^{circ}+theta)}{cos theta cos (60^{circ}-theta) cos (60^{circ}+theta)}`


`=frac{frac{1}{4} sin 3 theta}{frac{1}{4} cos 3 theta}=tan 3 theta`


Put `theta=20^{circ}`


So `tan theta tan (60^{circ}-theta) tan (60^{circ}+theta)=tan 20^{circ} tan 40^{circ} tan 80^{circ}=tan 3 times 20^{circ}=tan 60^{circ}=`


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