Inverse Trigonometric Functions Question 3
Question 3 - 30 January - Shift 2
Let $a_1=1, a_2, a_3, a_4, \ldots .$. be consecutive natural numbers. Then $\tan ^{-1}(\frac{1}{1+a_1 a_2})+\tan ^{-1}(\frac{1}{1+a_2 a_3})$ $+\ldots ….+\tan ^{-1}(\frac{1 \text{ thono }}{1+a _{2021} a _{2022}})$ is equal to
(1) $\frac{\pi}{4}-\cot ^{-1}(2022)$
(2) $\cot ^{-1}(2022)-\frac{\pi}{4}$
(3) $\tan ^{-1}(2022)-\frac{\pi}{4}$
(4) $\frac{\pi}{4}-\tan ^{-1}(2022)$
Show Answer
Answer: (3)
Solution:
Formula: Identity of subtraction of inverse tangent function.
$ \begin{aligned} & a_2-a_1=a_3-a_2=\ldots \ldots=a _{2022}-a _{2021}=1 \\ & \therefore \tan ^{-1}(\frac{a_2-a_1}{1+a_1 a_2})+\tan ^{-1}(\frac{a_3-a_2}{1+a_2 a_3})+\ldots . .+\tan ^{-1}(\frac{a _{2022}-a _{2021}}{1+a _{2021} a _{2022}}) \\ & =[(\tan ^{-1} a_2)-\tan ^{-1} a_1]+[\tan ^{-1} a_3-\tan ^{-1} a_2]+\ldots \ldots \\ & +[\tan ^{-1} a _{2022}-\tan ^{-1} a _{2021}] \\ & =\tan ^{-1} a _{2022}-\tan ^{-1} a_1 \\ & =\tan ^{-1}(2022)-\tan ^{-1} 1=\tan ^{-1} 2022-\frac{\pi}{4}(\text{ option 3) } \\ & =(\frac{\pi}{2}-\cot ^{-1}(2022))-\frac{\pi}{4} \\ & =\frac{\pi}{4}-\cot ^{-1}(2022) \text{ (option 1) } \end{aligned} $
