If tan px−tan qx=0,

Solution:

(a) AP

Given:

$\tan p x-\tan q x=0$

 

$\Rightarrow \tan p x=\tan q x$

$\Rightarrow \frac{\sin p x}{\cos p x}=\frac{\sin q x}{\cos q x}$

 

$\Rightarrow \sin p x \cos q x=\sin q x \cos p x$

$\Rightarrow \frac{1}{2}\left[\sin \left(\frac{p+q}{2}\right) x+\sin \left(\frac{p-q}{2}\right) x\right]=\frac{1}{2}\left[\sin \left(\frac{q+p}{2}\right) x+\sin \left(\frac{q-p}{2}\right) x\right]$

Now,

$\sin A \cos B=\frac{1}{2}\left[\sin \left(\frac{A+B}{2}\right)+\sin \left(\frac{A-B}{2}\right)\right]$

$\Rightarrow \sin \left(\frac{p-q}{2}\right) x=\sin \left(\frac{q-p}{2}\right) x$

$\Rightarrow \sin \left(\frac{p-q}{2}\right) x=-\sin \left(\frac{p-q}{2}\right) x$

$\Rightarrow 2 \sin \left(\frac{p-q}{2}\right) x=0$

$\Rightarrow \sin \left(\frac{p-q}{2}\right) x=0$

$\Rightarrow\left(\frac{p-q}{2}\right) x=n \pi, \mathrm{n} \in \mathrm{Z}$

$\Rightarrow x=\frac{2 \mathrm{n} \pi}{(\mathrm{p}-\mathrm{q})}, \mathrm{n} \in \mathrm{Z}$

Now, on putting the value of n">n, we get:

$n=1, x=\frac{2 \pi}{(p-q)}=a_{1}$

$n=2, \quad x=\frac{4 \pi}{(p-q)}=a_{2}$

$n=3, \quad x=\frac{6 \pi}{(p-q)}=a_{3}$

$n=4, x=\frac{8 \pi}{(p-q)}=a_{4}$

And so on.

Also,

$d=a_{2}-a_{1}=\frac{4 \pi}{(p-q)}-\frac{2 \pi}{(p-q)}=\frac{2 \pi}{(p-q)}$

$d=a_{3}-a_{2}=\frac{6 \pi}{(p-q)}-\frac{4 \pi}{(p-q)}=\frac{2 \pi}{(p-q)}$

$d=a_{4}-a_{3}=\frac{8 \pi}{(p-q)}-\frac{6 \pi}{(p-q)}=\frac{2 \pi}{(p-q)}$

And so on.

Thus,x forms a series in AP.