Let a_1=1, a_2, a_3, a_4, \ldots . . be consecutive natural numbers. Then \tan ^-1(\frac11+a_1 a_2)+\tan ^-1(\frac11+a_2 a_3)+\ldots . .+\tan ^-1(\frac11+a_2021 a_2022) is equal to :

\frac\pi4-\cot ^-1(2022), \tan ^-1(2022)-\frac\pi4

Inverse Trigonometric Functions PYQ — JEE Main 2023 · 30 Jan · Shift 2 · Q23 with Solution

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Correct answer: A, D

Identify the sequence

Since $a_1=1$ and $a_1,a_2,a_3,\dots$ are consecutive natural numbers, $a_n=n.$ So the sum becomes $S=\sum_{n=1}^{2021} \tan^{-1}\left(\frac{1}{1+n(n+1)}\right).$

Now, $1+n(n+1)=n^2+n+1.$ Hence $S=\sum_{n=1}^{2021} \tan^{-1}\left(\frac{1}{n^2+n+1}\right).$

Rewrite each term using arctangent subtraction

Recall: $\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\frac{x-y}{1+xy}\right)$ whenever the principal values match appropriately.

Take $x=n+1,\quad y=n.$ Then

=\tan^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right) =\tan^{-1}\left(\frac{1}{n^2+n+1}\right).$$ So each term is $$\tan^{-1}\left(\frac{1}{1+a_na_{n+1}}\right)=\tan^{-1}(n+1)-\tan^{-1}(n).$$ 3. **Telescoping the series** Therefore, $$S=\sum_{n=1}^{2021}\big[\tan^{-1}(n+1)-\tan^{-1}(n)\big].$$ This is telescoping: $$S=(\tan^{-1}2-\tan^{-1}1)+(\tan^{-1}3-\tan^{-1}2)+\cdots+(\tan^{-1}2022-\tan^{-1}2021).$$ All intermediate terms cancel, giving $$S=\tan^{-1}(2022)-\tan^{-1}(1).$$ Since $$\tan^{-1}(1)=\frac{\pi}{4},$$ we get $$S=\tan^{-1}(2022)-\frac{\pi}{4}.$$ 4. **Match with the options** So directly, $$S=\tan^{-1}(2022)-\frac{\pi}{4},$$ which is **Option D**. Now use the identity for positive $x$: $$\cot^{-1}(x)=\frac{\pi}{2}-\tan^{-1}(x).$$ Thus $$\frac{\pi}{4}-\cot^{-1}(2022) =\frac{\pi}{4}-\left(\frac{\pi}{2}-\tan^{-1}(2022)\right) =\tan^{-1}(2022)-\frac{\pi}{4}.$$ Hence **Option A** is also equal to the same value. So the correct options are: $$\boxed{A, D}.$$

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View written solutionFree

Correct answer: A, D

Identify the sequence

Since $a_1=1$ and $a_1,a_2,a_3,\dots$ are consecutive natural numbers, $a_n=n.$ So the sum becomes $S=\sum_{n=1}^{2021} \tan^{-1}\left(\frac{1}{1+n(n+1)}\right).$

Now, $1+n(n+1)=n^2+n+1.$ Hence $S=\sum_{n=1}^{2021} \tan^{-1}\left(\frac{1}{n^2+n+1}\right).$

Rewrite each term using arctangent subtraction

Recall: $\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\frac{x-y}{1+xy}\right)$ whenever the principal values match appropriately.

Take $x=n+1,\quad y=n.$ Then

=\tan^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right) =\tan^{-1}\left(\frac{1}{n^2+n+1}\right).$$ So each term is $$\tan^{-1}\left(\frac{1}{1+a_na_{n+1}}\right)=\tan^{-1}(n+1)-\tan^{-1}(n).$$ 3. **Telescoping the series** Therefore, $$S=\sum_{n=1}^{2021}\big[\tan^{-1}(n+1)-\tan^{-1}(n)\big].$$ This is telescoping: $$S=(\tan^{-1}2-\tan^{-1}1)+(\tan^{-1}3-\tan^{-1}2)+\cdots+(\tan^{-1}2022-\tan^{-1}2021).$$ All intermediate terms cancel, giving $$S=\tan^{-1}(2022)-\tan^{-1}(1).$$ Since $$\tan^{-1}(1)=\frac{\pi}{4},$$ we get $$S=\tan^{-1}(2022)-\frac{\pi}{4}.$$ 4. **Match with the options** So directly, $$S=\tan^{-1}(2022)-\frac{\pi}{4},$$ which is **Option D**. Now use the identity for positive $x$: $$\cot^{-1}(x)=\frac{\pi}{2}-\tan^{-1}(x).$$ Thus $$\frac{\pi}{4}-\cot^{-1}(2022) =\frac{\pi}{4}-\left(\frac{\pi}{2}-\tan^{-1}(2022)\right) =\tan^{-1}(2022)-\frac{\pi}{4}.$$ Hence **Option A** is also equal to the same value. So the correct options are: $$\boxed{A, D}.$$

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