JEE Main 2021 · 17 March · Morning Shift · Mathematics · Inverse Trigonometric Functions
A. 1.011 . 011.01
Given cot−1(α)=cot−1(2)+cot−1(8)+cot−1(18)+….\cot^{- 1} ( \alpha ) = \cot^{- 1} ( 2 ) + \cot^{- 1} ( 8 ) + \cot^{- 1} ( 18 ) + \ldots .cot−1(α)=cot−1(2)+cot−1(8)+cot−1(18)+….
=∑n=1100tan−1(24n2)= \mathop{∑}_{n = 1}^{100}\tan^{- 1}\left(\dfrac{2}{4{n}^{2}}\right)=∑n=1100tan−1(4n22)
=∑n=1100tan−1((2n+1)−(2n−1)1+(2n+1)(2n−1))= \mathop{∑}_{n = 1}^{100}\tan^{- 1}\left(\dfrac{( 2n + 1 ) - ( 2n - 1 )}{1 + ( 2n + 1 ) ( 2n - 1 )}\right)=∑n=1100tan−1(1+(2n+1)(2n−1)(2n+1)−(2n−1))
=∑n=1100tan−1(2n+1)−tan−1(2n−1)= \mathop{∑}_{n = 1}^{100}\tan^{- 1}\left(2n + 1\right) - \tan^{- 1}\left(2n - 1\right)=∑n=1100tan−1(2n+1)−tan−1(2n−1)
=tan−1201−tan−11= \tan^{- 1}201 - \tan^{- 1}1=tan−1201−tan−11
=tan−1(200202)= \tan^{- 1}\left(\dfrac{200}{202}\right)=tan−1(202200)
∴cot−1(α)=cot−1(202200)∴ \cot^{- 1} ( \alpha ) = \cot^{- 1}\left(\dfrac{202}{200}\right)∴cot−1(α)=cot−1(200202)
α=1.01\alpha = 1 . 01α=1.01
