The smallest positive value of x (in degrees) | vedclass

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(D) Given equation: $	an(x+100^{\circ}) = 	an(x+50^{\circ}) 	an(x) 	an(x-50^{\circ})$Using the identity $	an(A+B)	an(A-B) = \frac{\cos(2B)-\cos(

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$30$

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(D) Given equation: $	an(x+100^{\circ}) = 	an(x+50^{\circ}) 	an(x) 	an(x-50^{\circ})$Using the identity $	an(A+B)	an(A-B) = \frac{\cos(2B)-\cos(2A)}{\cos(2B)+\cos(2A)}$,we can rewrite the equation.Alternatively,using $	an(x+50^{\circ})	an(x-50^{\circ}) = \frac{\cos(100^{\circ})-\cos(2x)}{\cos(100^{\circ})+\cos(2x)}$.Let $A = x+50^{\circ}$ and $B = 50^{\circ}$. Then $	an(A+B)	an(A-B) = \frac{\cos(100^{\circ})-\cos(2x+100^{\circ})}{\cos(100^{\circ})+\cos(2x+100^{\circ})}$.Solving the equation leads to $\sin(4x+100^{\circ}) = -\cos(50^{\circ})$.$\sin(4x+100^{\circ}) = \sin(270^{\circ}-50^{\circ}) = \sin(220^{\circ})$.$4x+100^{\circ} = 220^{\circ}$ $\Rightarrow 4x = 120^{\circ}$ $\Rightarrow x = 30^{\circ}$.

The smallest positive value of $x$ (in degrees) for which $\tan(x+100^{\circ}) = \tan(x+50^{\circ}) \tan(x) \tan(x-50^{\circ})$ is (in $^{\circ}$)

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