- Exam
- JEE Advanced
- Year
- 2013
- Paper
- Paper 1
- Subject
- Mathematics
- Chapter
- Inverse Trigonometric Functions
JEE Advanced 2013 · Paper 1 · Mathematics · Inverse Trigonometric Functions
The value of cot ( Σ _ n = 1 ^ 23 cot ^ - 1 (1 + Σ _ k = 1 ^ n 2k))...
Options
- A
- B
- C
- D
Correct answer
B.
Step-by-step solution
\text{=cot}Σ{\text{cot}}^{- 1}\left(1 + \text{n}\left(\text{n} + 1\right)\right)\text{=cot}Σ{\text\tan}^{- 1}\dfrac{\left(\text{n} + 1\right) - \text{n}}{1 + \text{n}\left(\text{n} + 1\right)}
\text{=cot}\mathop{Σ}_{\text{n} = 1}^{23}\left({\text\tan}^{- 1}\left(\text{n} + 1\right) - {\text\tan}^{- 1}\text{n}\right)\left[{\text\tan}^{\text{-1}}\left(\dfrac{\text{x-y}}{1 + xy}\right){\text{=tan}}^{\text{-1}}{\text{x-tan}}^{\text{-1}}\text{y}\right]
Now {\text\tan}^{- 1}\left(\text{n} + 1\right) - {\text\tan}^{- 1}{\text{n=tan}}^{- 1}\left(2\right) - {\text\tan}^{- 1}{\text{1+tan}}^{- 1}\left(3\right) - {\text\tan}^{- 1}{\text{2+....+tan}}^{- 1}\left(24\right) - {\text\tan}^{- 1}\text{23}
\text{=cot}\left({\text\tan}^{- 1}24 - {\text\tan}^{- 1}1\right)
\text{=cot}\left({\text\tan}^{- 1}\dfrac{24 - 1}{1 + 24}\right)