The smallest positive value of x in degrees s | vedclass

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

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$30$

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(A) Given equation: $	an(x+100^{\circ}) = 	an(x+50^{\circ}) 	an(x) 	an(x-50^{\circ})$Using the identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we can simplify the expression.Alternatively,using $	an(x+100^{\circ}) = 	an(x+50^{\circ}+50^{\circ}) = \frac{	an(x+50^{\circ}) + 	an 50^{\circ}}{1 - 	an(x+50^{\circ}) 	an 50^{\circ}}$.By substituting $x = 30^{\circ}$:$LHS$: $	an(30^{\circ}+100^{\circ}) = 	an(130^{\circ}) = 	an(180^{\circ}-50^{\circ}) = -	an 50^{\circ}$.$RHS$: $	an(30^{\circ}+50^{\circ}) 	an(30^{\circ}) 	an(30^{\circ}-50^{\circ}) = 	an(80^{\circ}) 	an(30^{\circ}) 	an(-20^{\circ})$.Using the property $	an(3x) = 	an(x) 	an(60^{\circ}-x) 	an(60^{\circ}+x)$,we can see that for $x=30^{\circ}$,the equation holds true.Thus,the smallest positive value is $x = 30^{\circ}$.

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

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