Inverse Trigonometry Numerical JEE Main Previous Year Trigonometry Questions and Solutions - Solution of Triangles

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Solution

Q.
For $n \in N, if \cot^{- 1} 3 + \cot^{- 1} 4 + \cot^{- 1} 5 + \cot^{- 1} n = \frac{π}{4},$ then $n$ is equal to ______. [2024]

Ans.
(47)

$\cot^{- 1} 3 + \cot^{- 1} 4 + \cot^{- 1} 5 + \cot^{- 1} n = \frac{π}{4}$

$\Rightarrow \tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{4} + \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{n} = \frac{π}{4}$

$\Rightarrow \tan^{- 1} (\frac{\frac{1}{3} + \frac{1}{4}}{1 - \frac{1}{12}}) + \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{n} = \frac{π}{4}$

$\Rightarrow \tan^{- 1} (\frac{7}{11}) + \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{n} = \tan^{- 1} 1$

$\Rightarrow \tan^{- 1} (\frac{(\frac{7}{11} + \frac{1}{5})}{1 - \frac{7}{11} \cdot \frac{1}{5}}) + \tan^{- 1} \frac{1}{n} = \tan^{- 1} 1$

$\Rightarrow \tan^{- 1} (\frac{23}{24}) + \tan^{- 1} \frac{1}{n} = \tan^{- 1} 1$

$\Rightarrow \tan^{- 1} \frac{1}{n} = \tan^{- 1} 1 - \tan^{- 1} \frac{23}{24}$

$\Rightarrow \tan^{- 1} \frac{1}{n} = \tan^{- 1} (\frac{1 - \frac{23}{24}}{1 + \frac{23}{24}}) \Rightarrow \frac{1}{n} = \frac{1}{47} \Rightarrow n = 47$

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