will be in [Karnataka CET 1998]
A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: A
Solution :
We have \[\tan n\theta =\tan m\theta \]\[\Rightarrow \]\[n\theta =N\pi +(m\theta )\] \[\Rightarrow \] \[\theta =\frac{N\pi }{n-m}\], 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\[d=\frac{\pi }{n-m}\].
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