Let a_1=1,a_2,a_3,a_4,… .. be consecutive natural numbers. Then tan^-1( 1 1+a_1 a_2)+ tan^-1( 1 1+a_2 a_3) +… ..+ tan^-1( 1 1+a_2021 a_2022) is equal…

Given,

$${\mathrm{tan}}^{-1}(\frac{1}{1+{a}_{1}{a}_{2}})+{\mathrm{tan}}^{-1}(\frac{1}{1+{a}_{2}{a}_{3}})................{\mathrm{tan}}^{-1}(\frac{1}{1+{a}_{2021}{a}_{2022}})$$tan−1(1+a1​a2​1​)+tan−1(1+a2​a3​1​)................tan−1(1+a2021​a2022​1​)

$$={\mathrm{tan}}^{-1}(\frac{{a}_{2}-{a}_{1}}{1+{a}_{1}{a}_{2}})+{\mathrm{tan}}^{-1}(\frac{{a}_{3}-{a}_{2}}{1+{a}_{2}{a}_{3}})................{\mathrm{tan}}^{-1}(\frac{{a}_{2022}-{a}_{2021}}{1+{a}_{2021}{a}_{2022}})$$=tan−1(1+a1​a2​a2​−a1​​)+tan−1(1+a2​a3​a3​−a2​​)................tan−1(1+a2021​a2022​a2022​−a2021​​)

$$={\mathrm{tan}}^{-1}{a}_{2}-{\mathrm{tan}}^{-1}{a}_{1}+{\mathrm{tan}}^{-1}{a}_{3}-{\mathrm{tan}}^{-1}{a}_{2}..........+{\mathrm{tan}}^{-1}{a}_{2022}-{\mathrm{tan}}^{-1}{a}_{2021}$$=tan−1a2​−tan−1a1​+tan−1a3​−tan−1a2​..........+tan−1a2022​−tan−1a2021​

$${\text{using the formula }{\mathrm{tan}}^{-1}x-{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}(\frac{x-y}{1+xy})}$$using the formula tan−1x−tan−1y=tan−1(1+xyx−y​)

$$={\mathrm{tan}}^{-1}{a}_{2022}-{\mathrm{tan}}^{-1}{a}_{1}$$=tan−1a2022​−tan−1a1​

$$={\mathrm{tan}}^{-1}{a}_{2022}-{\mathrm{tan}}^{-1}1{\text{as given}{a}_{1}=1}$$=tan−1a2022​−tan−11as givena1​=1

$$={\mathrm{tan}}^{-1}{a}_{2022}-\frac{\pi }{4}$$=tan−1a2022​−4π​

$$=\frac{\pi }{2}-{\mathrm{cot}}^{-1}{a}_{2022}-\frac{\pi }{4}$$=2π​−cot−1a2022​−4π​

$$=\frac{\pi }{4}-{\mathrm{cot}}^{-1}{a}_{2022}$$=4π​−cot−1a2022​

$$=\frac{\pi }{4}-{\mathrm{cot}}^{-1}2022$$=4π​−cot−12022

As ${a}_{1}=1$a1​=1, ${a}_{2}=2$a2​=2 .... and so on ${a}_{2022}=2022$a2022​=2022