Question

The value of ${\tan }^{-1}\frac{4}{7}+{\tan }^{-1}\frac{4}{19}+{\tan }^{-1}\frac{4}{39}+{\tan }^{-1}\frac{4}{67}+\cdots \infty$ equals

• A. ${\tan }^{-1}1+{\tan }^{-1}\frac{1}{2}$
• B. $\frac{\pi }{2}-{\cot }^{-1}2$
• C. $\frac{\pi }{2}-{\cot }^{-1}1$
• D. ${\cot }^{-1}1+{\tan }^{-1}3$

Answer 1

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

Let $S=7+19+39+67+\cdots +T_n$
$S=0+7+19+39+67+\cdots +T_{n-1}+T_n$
Subtracting the above equations, we get
$T_n=7+12+20+28+\cdots +(T_n-T_{n-1})$
$=7+\frac{n-1}{2}[24+8(n-2\left)\right]$
$=4n^2+3$

Now, $T_n^{\prime }={\tan }^{-1}\left(\frac{4}{4n^2+3}\right)={\tan }^{-1}\left(\frac{1}{n^2+\frac{3}{4}}\right)$
$={\tan }^{-1}\frac{1}{1+(n^2-\frac{1}{4})}$
$={\tan }^{-1}\left[\frac{(n+\frac{1}{2})-(n-\frac{1}{2})}{1+(n+\frac{1}{2})(n-\frac{1}{2})}\right]$
$={\tan }^{-1}(n+\frac{1}{2})-{\tan }^{-1}(n-\frac{1}{2})$

Hence, $S_{\infty }=\sum _{n=1}^{\infty }{T}_n^{\prime }=\frac{\pi }{2}-{\tan }^{-1}\frac{1}{2}$
$=\frac{\pi }{2}-{\cot }^{-1}2$