Question

$\stackrel{Lim}{n\to \infty }[tan\theta +\frac{1}{2}tan\frac{\theta }{2}+\cdots \cdots \cdots +\frac{1}{2^n}tan\frac{\theta }{2^n}]=$

• A. $\frac{1}{\theta }$
• B. $\frac{1}{\theta }-2cot\theta$
• C. $2cot2\theta$
• D. $3$

Answer 1

The correct option is B $\frac{1}{\theta }-2cot\theta$

$cot\alpha -tan\alpha =2cot2\alpha$
$\stackrel{Lt}{n\to \infty }(-2cot2\theta +\frac{1}{2^n}cot\frac{\theta }{2^n})=-2cot2\theta +\stackrel{Lt}{n\to \infty }\frac{\frac{1}{2^n}}{tan\left(\frac{\theta }{2^n}\right)}=-2cot2\theta +\frac{1}{\theta }$