Question

The value of $ta{n}^{-1}\frac{4}{7}+ta{n}^{-1}\frac{4}{19}+ta{n}^{-1}\frac{4}{39}+ta{n}^{-1}\frac{4}{67}+...\infty$ equals

• A. $ta{n}^{-1}1+ta{n}^{-1}\frac{1}{2}+ta{n}^{-1}\frac{1}{3}$
• B. $ta{n}^{-1}1+co{t}^{-1}3$
• C. $co{t}^{-1}1+co{t}^{-1}\frac{1}{2}+co{t}^{-1}\frac{1}{3}$
• D. $co{t}^{-1}1+ta{n}^{-1}3$

Answer 1

Answer: (B)

$ta{n}^{-1}1+co{t}^{-1}3$

Let, $S_n={\tan }^{-1}\frac{4}{7}+{\tan }^{-1}\frac{4}{19}+{\tan }^{-1}\frac{4}{39}+{\tan }^{-1}\frac{4}{67}+..........\infty$

$S_n={lim}_{n\to \infty }\left({\sum }_{r=1}^n{\tan }^{-1}\left(\frac{4}{4r^2+3}\right)\right)$

$={lim}_{n\to \infty }{\sum }_{r=1}^n({\tan }^{-1}\left((r+\frac{1}{2})\right)-{\tan }^{-1}(r-\frac{1}{2}))$

$={\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(0\right)+{\tan }^{-1}\left(\frac{3}{4}\right)$

$={\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{3}{4}\right)$$={\tan }^{-1}\left(2\right)$

And ${\tan }^{-1}\left(1\right)+{\cot }^{-1}\left(3\right)={\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(\frac{1}{3}\right)={\tan }^{-1}\left(2\right)$