Question

Let $a_1=1, a_2, a_3, a_4, \ldots$ be consecutive natural numbersThen $\tan ^{-1}\left(\frac{1}{1+a_1 a_2}\right)+\tan ^{-1}\left(\frac{1}{1+a_2 a_3}\right)+\ldots +\tan ^{-1}\left(\frac{1}{1+a_{2021} a_{2022}}\right)$ is equal to

• A. $\cot ^{-1}(2022)-\frac{\pi}{4}$
• B. $\frac{\pi}{4}-\cot ^{-1}(2022)$
• C. $\tan ^{-1}(2022)-\frac{\pi}{4}$
• D. $\frac{\pi}{4}-\tan ^{-1}(2022)$

Answer 1

Answer: (B)

$\frac{\pi}{4}-\cot ^{-1}(2022)$

Answer 2

a2​−a1​=a3​−a2​=…..=a2022​−a2021​=1.
∴tan−1(1+a1​a2​a2​−a1​​)+tan−1(1+a2​a3​a3​−a2​​)+…..+tan−1(1+a2021​a2022​a2022​−a2021​​)
=[(tan−1a2​)−tan−1a1​]+[tan−1a3​−tan−1a2​]+…..+[tan−1a2022​−tan−1a2021​]
=tan−1a2022​−tan−1a1​
=tan−1(2022)−tan−11=tan−12022−4π​ (option 3)
=(2π​−cot−1(2022))−4π​
=4π​−cot−1(2022)( option 2)
So, the correct option is (B) : $\frac{\pi}{4}-\cot ^{-1}(2022)$