If \sum\limits_r = 1^50 \tan ^ - 11 \over 2r^2 = p , then the value of tan p is :

Inverse Trigonometric Functions PYQ — JEE Main 2021 · 26 Aug · Shift 2 · Q30 with Solution

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Correct answer: B

We need to evaluate $p=\sum_{r=1}^{50}\tan^{-1}\left(\frac{1}{2r^2}\right)$ and then find $\tan p$ .
Use the identity $\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\frac{x-y}{1+xy}\right)$ when the angles are in the principal range.

We try to write $\frac{1}{2r^2} = \frac{\frac{1}{2r-1}-\frac{1}{2r+1}}{1+\frac{1}{(2r-1)(2r+1)}}.$ Let us verify it:

=\frac{\frac{(2r+1)-(2r-1)}{(2r-1)(2r+1)}}{\frac{(2r-1)(2r+1)+1}{(2r-1)(2r+1)}}$$ $$=\frac{\frac{2}{(2r-1)(2r+1)}}{\frac{4r^2}{(2r-1)(2r+1)}} =\frac{2}{4r^2} =\frac{1}{2r^2}.$$ So, $$\tan^{-1}\left(\frac{1}{2r^2}\right)=\tan^{-1}\left(\frac{1}{2r-1}\right)-\tan^{-1}\left(\frac{1}{2r+1}\right).$$ 3. Therefore the sum telescopes: $$p=\sum_{r=1}^{50}\left[\tan^{-1}\left(\frac{1}{2r-1}\right)-\tan^{-1}\left(\frac{1}{2r+1}\right)\right].$$ Write first few terms: $$p=\left(\tan^{-1}1-\tan^{-1}\frac13\right)+\left(\tan^{-1}\frac13-\tan^{-1}\frac15\right)+\cdots+\left(\tan^{-1}\frac{1}{99}-\tan^{-1}\frac{1}{101}\right).$$ Everything cancels except the first and last terms: $$p=\tan^{-1}(1)-\tan^{-1}\left(\frac{1}{101}\right).$$ 4. Now compute $\tan p$: $$\tan p=\tan\left(\tan^{-1}(1)-\tan^{-1}\left(\frac{1}{101}\right)\right).$$ Using $$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B},$$ we get $$\tan p=\frac{1-\frac{1}{101}}{1+1\cdot \frac{1}{101}}$$ $$=\frac{\frac{100}{101}}{\frac{102}{101}}=\frac{100}{102}=\frac{50}{51}.$$ 5. Hence the correct option is $$\boxed{\frac{50}{51}}.$$ So, option **B** is correct.

View written solutionFree

Correct answer: B

We need to evaluate $p=\sum_{r=1}^{50}\tan^{-1}\left(\frac{1}{2r^2}\right)$ and then find $\tan p$ .
Use the identity $\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\frac{x-y}{1+xy}\right)$ when the angles are in the principal range.

We try to write $\frac{1}{2r^2} = \frac{\frac{1}{2r-1}-\frac{1}{2r+1}}{1+\frac{1}{(2r-1)(2r+1)}}.$ Let us verify it:

=\frac{\frac{(2r+1)-(2r-1)}{(2r-1)(2r+1)}}{\frac{(2r-1)(2r+1)+1}{(2r-1)(2r+1)}}$$ $$=\frac{\frac{2}{(2r-1)(2r+1)}}{\frac{4r^2}{(2r-1)(2r+1)}} =\frac{2}{4r^2} =\frac{1}{2r^2}.$$ So, $$\tan^{-1}\left(\frac{1}{2r^2}\right)=\tan^{-1}\left(\frac{1}{2r-1}\right)-\tan^{-1}\left(\frac{1}{2r+1}\right).$$ 3. Therefore the sum telescopes: $$p=\sum_{r=1}^{50}\left[\tan^{-1}\left(\frac{1}{2r-1}\right)-\tan^{-1}\left(\frac{1}{2r+1}\right)\right].$$ Write first few terms: $$p=\left(\tan^{-1}1-\tan^{-1}\frac13\right)+\left(\tan^{-1}\frac13-\tan^{-1}\frac15\right)+\cdots+\left(\tan^{-1}\frac{1}{99}-\tan^{-1}\frac{1}{101}\right).$$ Everything cancels except the first and last terms: $$p=\tan^{-1}(1)-\tan^{-1}\left(\frac{1}{101}\right).$$ 4. Now compute $\tan p$: $$\tan p=\tan\left(\tan^{-1}(1)-\tan^{-1}\left(\frac{1}{101}\right)\right).$$ Using $$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B},$$ we get $$\tan p=\frac{1-\frac{1}{101}}{1+1\cdot \frac{1}{101}}$$ $$=\frac{\frac{100}{101}}{\frac{102}{101}}=\frac{100}{102}=\frac{50}{51}.$$ 5. Hence the correct option is $$\boxed{\frac{50}{51}}.$$ So, option **B** is correct.