The number of elements in the set {x in [0, 1 | vedclass

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(C) Given equation: $	an(x+100^{\circ}) = 	an(x+50^{\circ}) 	an x 	an(x-50^{\circ})$Using $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,

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$4$

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(C) Given equation: $	an(x+100^{\circ}) = 	an(x+50^{\circ}) 	an x 	an(x-50^{\circ})$Using $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we note that $	an(x+50^{\circ}) 	an(x-50^{\circ}) = \frac{	an^2 x - 	an^2 50^{\circ}}{1 - 	an^2 x 	an^2 50^{\circ}}$.Alternatively,using the identity $	an(A+B)	an(A-B) = \frac{\sin^2 A - \sin^2 B}{\cos^2 A - \cos^2 B} = \frac{	an^2 A - 	an^2 B}{1 - 	an^2 A 	an^2 B}$.After simplifying the trigonometric expression,we get $\sin(4x + 100^{\circ}) = \sin(-40^{\circ})$.General solution: $4x + 100^{\circ} = n \cdot 180^{\circ} + (-1)^n (-40^{\circ})$.For $n=0, 4x = -140^{\circ} \implies x = -35^{\circ}$ (not in range).For $n=1, 4x = 180^{\circ} + 40^{\circ} - 100^{\circ} = 120^{\circ} \implies x = 30^{\circ}$.For $n=2, 4x = 360^{\circ} - 40^{\circ} - 100^{\circ} = 220^{\circ} \implies x = 55^{\circ}$.For $n=3, 4x = 540^{\circ} + 40^{\circ} - 100^{\circ} = 480^{\circ} \implies x = 120^{\circ}$.For $n=4, 4x = 720^{\circ} - 40^{\circ} - 100^{\circ} = 580^{\circ} \implies x = 145^{\circ}$.Thus,there are $4$ solutions in the interval $[0, 180^{\circ}]$.

The number of elements in the set $\{x \in [0, 180^{\circ}] : \tan(x+100^{\circ}) = \tan(x+50^{\circ}) \tan x \tan(x-50^{\circ})\}$ is . . . . . . .

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