If tan ntheta = tan mtheta,then the different | vedclass

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(A) Given $	an n	heta = 	an m	heta$.The general solution for $	an x = 	an y$ is $x = N\pi + y$,where $N \in \mathbb{Z}$.Therefore,$n	heta = N\p

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$A.P.$

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(A) Given $	an n	heta = 	an m	heta$.The general solution for $	an x = 	an y$ is $x = N\pi + y$,where $N \in \mathbb{Z}$.Therefore,$n	heta = N\pi + m	heta$.Rearranging the terms,we get $(n - m)	heta = N\pi$.Thus,$	heta = \frac{N\pi}{n - m}$.For different integer values of $N = 1, 2, 3, \dots$,the values of $	heta$ are $\frac{\pi}{n - m}, \frac{2\pi}{n - m}, \frac{3\pi}{n - m}, \dots$.This sequence has a common difference $d = \frac{\pi}{n - m}$,which implies that the values are in $A.P.$

If $\tan n\theta = \tan m\theta$,then the different values of $\theta$ will be in

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