Question

If $\tan px-\tan qx=0$, then the values of θ form a series in
(a) AP
(b) GP
(c) HP
(d) none of these

Answer 1

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

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

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

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

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

$n=4,x=\frac{8\mathrm{\pi }}{(p-q)}$ = 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, $x$ forms a series in AP.