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

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(C) Given,$	an(	heta+100^{\circ})=	an(	heta+50^{\circ}) 	an(	heta) 	an(	heta-50^{\circ})$.Using the identity $	an(A+B) = \frac{	an A + 	an

Vedclass · Accepted Answer

$30$

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(C) Given,$	an(	heta+100^{\circ})=	an(	heta+50^{\circ}) 	an(	heta) 	an(	heta-50^{\circ})$.Using the identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ and $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we can simplify the expression.Alternatively,using $	an(x) = \frac{\sin x}{\cos x}$,the equation becomes $\frac{\sin(	heta+100^{\circ})}{\cos(	heta+100^{\circ})} = \frac{\sin(	heta+50^{\circ}) \sin(	heta) \sin(	heta-50^{\circ})}{\cos(	heta+50^{\circ}) \cos(	heta) \cos(	heta-50^{\circ})}$.Applying componendo and dividendo,we get $\frac{\sin(	heta+100^{\circ})\cos(	heta-50^{\circ}) + \cos(	heta+100^{\circ})\sin(	heta-50^{\circ})}{\sin(	heta+100^{\circ})\cos(	heta-50^{\circ}) - \cos(	heta+100^{\circ})\sin(	heta-50^{\circ})} = \frac{\sin(	heta+50^{\circ})\sin(	heta) + \cos(	heta+50^{\circ})\cos(	heta)}{\sin(	heta+50^{\circ})\sin(	heta) - \cos(	heta+50^{\circ})\cos(	heta)}$.This simplifies to $\frac{\sin(2	heta+50^{\circ})}{\sin(150^{\circ})} = \frac{\cos(50^{\circ})}{-\cos(2	heta+50^{\circ})}$.$\Rightarrow -\sin(2	heta+50^{\circ})\cos(2	heta+50^{\circ}) = \sin(150^{\circ})\cos(50^{\circ})$.$\Rightarrow -\frac{1}{2}\sin(4	heta+100^{\circ}) = \frac{1}{2}\cos(50^{\circ})$.$\Rightarrow \sin(4	heta+100^{\circ}) = -\cos(50^{\circ}) = \sin(220^{\circ})$.$4	heta+100^{\circ} = 220^{\circ}$ $\Rightarrow 4	heta = 120^{\circ}$ $\Rightarrow 	heta = 30^{\circ}$.

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

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