Question

Find the value of the $\sum _{r=1}^{100}\frac{\tan \left(2^{r-1}\right)}{\cos 2^r}$ is

• A. $\tan 2^{99}-\tan 1$
• B. $\tan 2^{100}$
• C. $\tan 2^{100}-\tan 1$
• D. $\tan 2^{99}$

Answer 1

The correct option is C $\tan 2^{100}-\tan 1$

$T_r=\frac{\tan \left(2^{r-1}\right)}{\cos 2^r}$
Let $2^{r-1}=\theta$
$T=\frac{\tan \theta }{\cos 2\theta }=\frac{\sin \theta }{\cos \theta \cos 2\theta }\Rightarrow T=\frac{\sin (2\theta -\theta )}{\cos \theta \cos 2\theta }\Rightarrow T=\frac{\sin 2\theta \cos \theta -\sin \theta \cos 2\theta }{\cos \theta \cos 2\theta }\Rightarrow T=\tan 2\theta -\tan \theta \Rightarrow T_r=\tan 2^r-\tan 2^{r-1}$

$S=\sum _{r=1}^{100}\frac{\tan \left(2^{r-1}\right)}{\cos 2^r}=\sum _{r=1}^{100}{T}_r\Rightarrow S=\sum _{r=1}^{100}\tan 2^r-\tan 2^{r-1}\Rightarrow S=\tan 2-\tan 1+\tan 2^2-\tan 2..+\tan 2^{100}-\tan 2^{99}\Rightarrow S=\tan 2^{100}-\tan 1$