Question

The sum of ${\cot }^{-1}\frac{3}{14}+{\cot }^{-1}8+{\cot }^{-1}18+{\cot }^{-1}32=$___________ is?

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{12}$

Answer 1

The correct option is B $\frac{\pi }{2}$

We have,

${\cot }^{-1}\frac{3}{14}+{\cot }^{-1}8+{\cot }^{-1}18+{\cot }^{-1}32$

$\Rightarrow {\cot }^{-1}\frac{3}{14}+{\cot }^{-1}18+{\cot }^{-1}8+{\cot }^{-1}32$

We know that

${\cot }^{-1}x+{\cot }^{-1}y={\cot }^{-1}\frac{(xy-1)}{x+y}$

Therefore,

$\Rightarrow {\cot }^{-1}\left(\frac{\frac{3}{14}\times 18-1}{3/14+18}\right)+{\cot }^{-1}\left(\frac{8\times 32-1}{8+32}\right)$

$\Rightarrow {\cot }^{-1}\left(\frac{40}{255}\right)+{\cot }^{-1}\left(\frac{255}{40}\right)$

$\Rightarrow {\cot }^{-1}\left(0\right)$

$\Rightarrow {\cot }^{-1}\cot \frac{\pi }{2}$

$\Rightarrow \frac{\pi }{2}$

Hence, this is the answer.