The value of tan 20^{circ} tan 80^{circ} cot | vedclass

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Question

(A) Let $x = 	an 20^{\circ} 	an 80^{\circ} \cot 50^{\circ}$.We know that $\cot 50^{\circ} = \frac{1}{	an 50^{\circ}}$.So,$x = \frac{	an 20^{\circ}

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(A) Let $x = 	an 20^{\circ} 	an 80^{\circ} \cot 50^{\circ}$.We know that $\cot 50^{\circ} = \frac{1}{	an 50^{\circ}}$.So,$x = \frac{	an 20^{\circ} 	an 80^{\circ}}{	an 50^{\circ}}$.Using the identity $	an 	heta 	an(60^{\circ} - 	heta) 	an(60^{\circ} + 	heta) = 	an 3	heta$,we have $	an 20^{\circ} 	an(60^{\circ} - 20^{\circ}) 	an(60^{\circ} + 20^{\circ}) = 	an(3 	imes 20^{\circ}) = 	an 60^{\circ} = \sqrt{3}$.Thus,$	an 20^{\circ} 	an 40^{\circ} 	an 80^{\circ} = \sqrt{3}$.We need to evaluate $\frac{	an 20^{\circ} 	an 80^{\circ}}{	an 50^{\circ}}$.Using the identity $	an(60^{\circ} - 	heta) 	an 	heta 	an(60^{\circ} + 	heta) = 	an 3	heta$,we can write $	an 80^{\circ} = 	an(60^{\circ} + 20^{\circ})$.Also,$	an 50^{\circ} = 	an(60^{\circ} - 10^{\circ})$ is not directly helpful,but we know $	an 50^{\circ} = 	an(60^{\circ} - 10^{\circ})$ and $	an 70^{\circ} = 	an(60^{\circ} + 10^{\circ})$.Alternatively,$x = \frac{\sin 20^{\circ} \sin 80^{\circ} \cos 50^{\circ}}{\cos 20^{\circ} \cos 80^{\circ} \sin 50^{\circ}}$.Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$,we get $\sin 20^{\circ} \sin 80^{\circ} = \frac{1}{2}(\cos 60^{\circ} - \cos 100^{\circ}) = \frac{1}{2}(\frac{1}{2} + \sin 10^{\circ})$.After simplification,the value is $\sqrt{3}$.

The value of $\tan 20^{\circ} \tan 80^{\circ} \cot 50^{\circ} = $

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