Question

$Iftanp\theta -tanq\theta =0,thenthevaluesof\theta formaseriesin$

• A. AP
• B. GP
• C. HP
• D. none of these

Answer 1

The correct option is A

AP

$Given:tapp\theta -tanq\theta =0\Rightarrow tapp\theta -tanq\theta =0\Rightarrow \frac{sinp\theta }{cosp\theta }=\frac{sinq\theta }{cosq\theta }\Rightarrow sinp\theta cosq\theta =sinq\theta cosp\theta =0\Rightarrow \frac{1}{2}\left[\begin{array}{c}sin\left(\frac{p+q}{2}\right)\theta +sin\left(\frac{p-q}{2}\right)\theta \end{array}\right]=\frac{1}{2}\left[\begin{array}{c}sin\left(\frac{q+p}{2}\right)\theta +sin\left(\frac{q-p}{2}\right)\theta \end{array}\right]Now,sinAcosB=\frac{1}{2}\left[\begin{array}{c}sin\left(\frac{A+B}{2}\right)+sin\left(\frac{A-B}{2}\right)\end{array}\right]\Rightarrow sin\left(\frac{p-q}{2}\right)\theta =sin\left(\frac{q-p}{2}\right)\theta \Rightarrow sin\left(\frac{p-q}{2}\right)\theta =-sin\left(\frac{p-q}{2}\right)\theta \Rightarrow 2sin\left(\frac{p-q}{2}\right)\theta =0\Rightarrow sin\left(\frac{p-q}{2}\right)\theta =0\Rightarrow \left(\frac{p-q}{2}\right)\theta =n\pi ,n\in Z\text{Now,on putting the value of n,we get}n=1,\theta =\frac{2\pi }{(p-q)}=a_{1}n=2,\theta =\frac{4\pi }{(p-q)}=a_{2}n=3,\theta =\frac{6\pi }{(p-q)}=a_{3}n=4,\theta =\frac{8\pi }{(p-q)}=a_{4}Andsoon.Also,d=a_2-a_1=\frac{4\pi }{(p-q)}-\frac{2\pi }{(p-q)}=\frac{2\pi }{(p-q)}d=a_3-a_2=\frac{6\pi }{(p-q)}-\frac{4\pi }{(p-q)}=\frac{2\pi }{(p-q)}d=a_4-a_3=\frac{8\pi }{(p-q)}-\frac{6\pi }{(p-q)}=\frac{2\pi }{(p-q)}Andsoon.Thus,\theta formsaseriesinAP.$