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

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Question

(a) We have $	an n	heta = 	an m	heta $

$ \Rightarrow $$n	heta = N\pi + (m	heta )$

$ \Rightarrow $ $	heta = \frac{{N\pi }}{{n - m}}$,

p

Vedclass · Accepted Answer

$A.P.$

Vedclass · Answer

(a) We have $	an n	heta = 	an m	heta $

$ \Rightarrow $$n	heta = N\pi + (m	heta )$

$ \Rightarrow $ $	heta = \frac{{N\pi }}{{n - m}}$,

putting $N = 1,\;2,\;3.........,$ we get

$\frac{\pi }{{n - m}},\;\frac{{2\pi }}{{n - m}},\;\frac{{3\pi }}{{n - m}}.........$

which are obviously in $A.P.$

Since common difference $d = \frac{\pi }{{n - m}}$.

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

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