Question

If $\tan p\mathrm{\theta }-\tan q\mathrm{\theta }=0$, then the values of θ form a series in
(a) AP
(b) GP
(c) HP
(d) none of these

Answer 1

(a) AP
Given:
$\tan p\theta -\tan q\theta =0$
$$\begin{gathered} \Rightarrow \tan p\theta =\tan q\theta \\ \Rightarrow \frac{\sin p\theta }{\cos p\theta }=\frac{\sin q\theta }{\cos q\theta } \\ \Rightarrow \sin p\theta \cos q\theta =\sin q\theta \cos p\theta \\ \Rightarrow \frac{1}{2}\left[\sin \left(\frac{p+q}{2}\right)\theta +\sin \left(\frac{p-q}{2}\right)\theta \right]=\frac{1}{2}\left[\sin \left(\frac{q+p}{2}\right)\theta +\sin \left(\frac{q-p}{2}\right)\theta \right] \end{gathered}$$

Now,
$\sin A\cos B=\frac{1}{2}\left[\sin \left(\frac{A+B}{2}\right)+\sin \left(\frac{A-B}{2}\right)\right]$
$$\begin{gathered} \Rightarrow \sin \left(\frac{p-q}{2}\right)\theta =\sin \left(\frac{q-p}{2}\right)\theta \\ \Rightarrow \sin \left(\frac{p-q}{2}\right)\theta =-\sin \left(\frac{p-q}{2}\right)\theta \\ \Rightarrow 2\sin \left(\frac{p-q}{2}\right)\theta =0 \\ \Rightarrow \sin \left(\frac{p-q}{2}\right)\theta =0 \end{gathered}$$
$$\begin{gathered} \Rightarrow \left(\frac{p-q}{2}\right)\theta =n\mathrm{\pi },\mathrm{n}\in \mathrm{Z} \\ \Rightarrow \mathrm{\theta }=\frac{2\mathrm{n\pi }}{(\mathrm{p}-\mathrm{q})},\mathrm{n}\in \mathrm{Z} \end{gathered}$$

Now, on putting the value of $n$, we get:
$n=1,\theta =\frac{2\mathrm{\pi }}{(p-q)}$= a₁

$n=2,\theta =\frac{4\mathrm{\pi }}{(p-q)}$ = a₂

$n=3,\theta =\frac{6\mathrm{\pi }}{(p-q)}$ = a₃

$n=4,\theta =\frac{8\mathrm{\pi }}{(p-q)}$ = a₄
And so on.


Also,

$$\begin{gathered} d=a_2-a_1=\frac{4\mathrm{\pi }}{(p-q)}-\frac{2\mathrm{\pi }}{(p-q)}=\frac{2\mathrm{\pi }}{(p-q)} \\ d=a_3-a_2=\frac{6\mathrm{\pi }}{(p-q)}-\frac{4\mathrm{\pi }}{(p-q)}=\frac{2\mathrm{\pi }}{(p-q)} \\ d=a_4-a_3=\frac{8\mathrm{\pi }}{(p-q)}-\frac{6\mathrm{\pi }}{(p-q)}=\frac{2\mathrm{\pi }}{(p-q)} \end{gathered}$$
And so on.

Thus, $\theta$ forms a series in AP.