Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

$\begin{array}{rl}& {\cot }^{-1}3+{\cot }^{-1}4+{\cot }^{-1}5+{\cot }^{1}n=\frac{\pi }{4}\\ & {\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{4}+{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{\mathrm{n}}=\frac{\pi }{4}\\ & {\tan }^{-1}\left(\frac{46}{48}\right)+{\tan }^{-1}\frac{1}{\mathrm{n}}=\frac{\pi }{4}\\ & {\tan }^{-1}\left(\frac{23}{24}\right)+{\tan }^{-1}\frac{1}{\mathrm{n}}=\frac{\pi }{4}\\ & {\tan }^{-1}\frac{1}{\mathrm{n}}={\tan }^{-1}1-{\tan }^{-1}\frac{23}{24}\end{array}$
$\begin{array}{rl}& {\tan }^{-1}\frac{1}{n}={\tan }^{-1}1-{\tan }^{-1}\frac{23}{24}\\ & {\tan }^{-1}\frac{1}{n}={\tan }^{-1}\left(\frac{1-\frac{23}{24}}{1+\frac{23}{24}}\right)\\ & {\tan }^{-1}\frac{1}{n}={\tan }^{-1}\left(\frac{\frac{1}{24}}{\frac{47}{24}}\right)\\ & {\tan }^{-1}\frac{1}{n}={\tan }^{-1}\frac{1}{47}\\ & n=47\end{array}$