Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

The series can be written as ${\sum }_{n=1}^{\infty }{\cot }^{-1}(n^2+\frac{3}{4})={\sum }_{n=1}^{\infty }{\tan }^{-1}(\frac{1}{n^2+\frac{3}{4}})$.

We can rewrite the argument of the arctangent as $\frac{1}{n^2+\frac{3}{4}}=\frac{1}{(n+\frac{1}{2}{)}^2-n-\frac{1}{4}+\frac{3}{4}}=\frac{(n+\frac{1}{2})-(n-\frac{1}{2})}{1+(n+\frac{1}{2})(n-\frac{1}{2})}$.

Using the identity ${\tan }^{-1}(a)-{\tan }^{-1}(b)={\tan }^{-1}(\frac{a-b}{1+ab})$, the sum telescopes to ${\tan }^{-1}(n+\frac{1}{2}){|}_{n=1}^{\infty }-{\tan }^{-1}(n-\frac{1}{2}){|}_{n=1}^{\infty }={\lim }_{N\to \infty }{\tan }^{-1}(N+\frac{1}{2})-{\tan }^{-1}(\frac{1}{2})=\frac{\pi }{2}-{\tan }^{-1}(\frac{1}{2})={\tan }^{-1}(2)$.

Final Answer: The final answer is ${\tan }^{-1}2$