Question

If $\sum _{n=1}^{2013}\tan \left(\frac{\theta }{2^n}\right)\sec \left(\frac{\theta }{2^{n-1}}\right)=\tan \left(\frac{\theta }{2^a}\right)-\tan \left(\frac{\theta }{2^b}\right)$ then $(b+a)$ equals

• A. $2014$
• B. $2012$
• C. $2013$
• D. $2015$

Answer 1

Answer: (C)

$2013$

Let $\frac{\theta }{2^{n-1}}=A,\frac{\theta }{2^n}=\frac{A}{2}$

$\tan \frac{A}{2}\sec A=\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}\times \frac{1}{\cos A}$

$=\frac{\sin (A-\frac{A}{2})}{\cos \frac{A}{2}}\times \frac{1}{\cos A}$

$=\frac{(\sin A\cos \frac{A}{2}-\cos A\sin \frac{A}{2})\times \frac{1}{\cos A}}{\cos \frac{A}{2}}$

$=\tan A-\tan \frac{A}{2}$

$\sum _{n=1}^{2013}\tan \frac{\theta }{2^{n-1}}\sec \frac{\theta }{2^n}$

$f_{2013}\left(\theta \right)=\tan \theta -\tan \frac{\theta }{2}+\tan \frac{\theta }{2}-\tan \frac{\theta }{4}+\tan \frac{\theta }{4}-\tan \frac{\theta }{8}+....+2013$ terms

All the even places terms get canceled.

$=\frac{\tan \theta }{2^0}-\tan \frac{\theta }{2^{2013}}$

$\therefore a=0,b=2013$

Hence $b+a=2013+0=2013$