Question

If \[\tan px - \tan qx = 0\], then the values of θ form a series in

 

पर्याय

• AP

• GP

• HP

•  none of these

Answer 1

AP
Given:
\[\tan px - \tan qx = 0\]
\[\Rightarrow \tan px = \tan qx\]
\[ \Rightarrow \frac{\sin px}{\cos px} = \frac{\sin qx}{\cos qx}\]
\[ \Rightarrow \sin px \cos qx = \sin qx \cos px\]
\[ \Rightarrow \frac{1}{2}\left[ \sin\left( \frac{p + q}{2} \right)x + \sin\left( \frac{p - q}{2} \right)x \right] = \frac{1}{2}\left[ \sin\left( \frac{q + p}{2} \right)x + \sin\left( \frac{q - p}{2} \right)x \right]\]
Now,
\[\sin A \cos B = \frac{1}{2}\left[ \sin\left( \frac{A + B}{2} \right) + \sin\left( \frac{A - B}{2} \right) \right]\]
\[\Rightarrow \sin \left( \frac{p - q}{2} \right)x = \sin \left( \frac{q - p}{2} \right)x\]
\[ \Rightarrow \sin \left( \frac{p - q}{2} \right)x = - \sin \left( \frac{p - q}{2} \right)x\]
\[ \Rightarrow 2 \sin \left( \frac{p - q}{2} \right)x = 0\]
\[ \Rightarrow \sin \left( \frac{p - q}{2} \right)x = 0\]
\[\Rightarrow \left( \frac{p - q}{2} \right)x = n\pi, n \in Z\]
\[ \Rightarrow x = \frac{2n\pi}{(p - q)}, n \in Z\]
Now, on putting the value of 
n, we get: \[n = 1, x = \frac{2\pi}{(p - q)}\]= a₁

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

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

\[n = 4, x = \frac{8\pi}{(p - q)}\] = a₄

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.