For n ∈ N , if ^ -1 3+ ^ -1 4+ ^ -1 5+ ^ -1 n=(π)/(4) , then n is equal to_____

Correct answer: 0 aligned & ^ -1 3+ ^ -1 4+ ^ -1 5+ ^1 n=(π)/(4) & ^ -1 (1)/(3)+ ^ -1 (1)/(4)+ ^ -1 (1)/(5)+ ^ -1 1 n =(π)/(4) & ^ -1 ((46)/(48))+ ^ -1 1 n =(π)/(4) & ^ -1 ((23)/(24))+ ^ -1 1 n =(π)/(4) & ^ -1 1 n = ^ -1 1- ^ -1 (23)/(24) aligned aligned & ^ -1 (1)/(n)= ^ -1 1- ^ -1 (23)/(24) & ^ -1 (1)/(n)= ^ -1 ((1-(23)/(24))/(1+(23)/(24))) & ^ -1 (1)/(n)= ^ -1 (((1)/(24))/((47)/(24))) & ^ -1 (1)/(n)= ^ -1 (1)/(47) & n=47 aligned

JEE Main 20246 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 6 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

JEE Main 2024 · 6 April · Morning Shift · Mathematics · Inverse Trigonometric Functions

For n ∈ N , if ^ -1 3+ ^ -1 4+ ^ -1 5+ ^ -1 n=(π)/(4) , then n is...

For

n \in \mathrm{N}

, if

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

, then

n

is equal to_____

Correct answer

Step-by-step solution

\begin{aligned} & \cot ^{-1} 3+\cot ^{-1} 4+\cot ^{-1} 5+\cot ^1 n=\frac{\pi}{4} \\ & \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{\mathrm{n}}=\frac{\pi}{4} \\ & \tan ^{-1}\left(\frac{46}{48}\right)+\tan ^{-1} \frac{1}{\mathrm{n}}=\frac{\pi}{4} \\ & \tan ^{-1}\left(\frac{23}{24}\right)+\tan ^{-1} \frac{1}{\mathrm{n}}=\frac{\pi}{4} \\ & \tan ^{-1} \frac{1}{\mathrm{n}}=\tan ^{-1} 1-\tan ^{-1} \frac{23}{24}\end{aligned}

\begin{aligned} & \tan ^{-1} \frac{1}{n}=\tan ^{-1} 1-\tan ^{-1} \frac{23}{24} \\ & \tan ^{-1} \frac{1}{n}=\tan ^{-1}\left(\frac{1-\frac{23}{24}}{1+\frac{23}{24}}\right) \\ & \tan ^{-1} \frac{1}{n}=\tan ^{-1}\left(\frac{\frac{1}{24}}{\frac{47}{24}}\right) \\ & \tan ^{-1} \frac{1}{n}=\tan ^{-1} \frac{1}{47} \\ & n=47\end{aligned}

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