Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Let the general term be $T_n={\tan }^{-1}\left(\frac{2}{n^2+n-1}\right)$. We want to express $T_n$ in the form ${\tan }^{-1}(A)-{\tan }^{-1}(B)={\tan }^{-1}\left(\frac{A-B}{1+AB}\right)$.

We need $A-B=2$ and $1+AB=n^2+n-1$, which means $AB=n^2+n-2$. We look for $A$ and $B$ such that $A-B=2$ and $AB=(n+2)(n-1)$. Let $A=n+x$ and $B=n+y$. Then $(n+x)-(n+y)=x-y=2$. And $(n+x)(n+y)=n^2+(x+y)n+xy=n^2+n-2$. Comparing coefficients, $x+y=1$ and $xy=-2$. From $x-y=2$ and $x+y=1$, we get $2x=3⟹x=3/2$ and $y=-1/2$. So $A=n+3/2$ and $B=n-1/2$. Thus, $T_n={\tan }^{-1}\left(n+\frac{3}{2}\right)-{\tan }^{-1}\left(n-\frac{1}{2}\right)$. This is a telescopic series. Let $S_N={\sum }_{n=1}^N{T}_n$.$$S_N=\left({\tan }^{-1}\left(\frac{5}{2}\right)-{\tan }^{-1}\left(\frac{1}{2}\right)\right)+\left({\tan }^{-1}\left(\frac{7}{2}\right)-{\tan }^{-1}\left(\frac{3}{2}\right)\right)+\left({\tan }^{-1}\left(\frac{9}{2}\right)-{\tan }^{-1}\left(\frac{5}{2}\right)\right)+\cdots +\left({\tan }^{-1}\left(N+\frac{3}{2}\right)-{\tan }^{-1}\left(N-\frac{1}{2}\right)\right)$$ Most terms cancel out. We are left with:$$S_N={\tan }^{-1}\left(N+\frac{3}{2}\right)+{\tan }^{-1}\left(N+\frac{1}{2}\right)-{\tan }^{-1}\left(\frac{1}{2}\right)-{\tan }^{-1}\left(\frac{3}{2}\right)$$ As $N\to \infty$, ${\tan }^{-1}\left(N+\frac{3}{2}\right)\to \frac{\pi }{2}$ and ${\tan }^{-1}\left(N+\frac{1}{2}\right)\to \frac{\pi }{2}$. So the sum is $\frac{\pi }{2}+\frac{\pi }{2}-\left({\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{3}{2}\right)\right)$.$$=\pi -{\tan }^{-1}\left(\frac{\frac{1}{2}+\frac{3}{2}}{1-\frac{1}{2}\cdot \frac{3}{2}}\right)=\pi -{\tan }^{-1}\left(\frac{2}{1-\frac{3}{4}}\right)=\pi -{\tan }^{-1}\left(\frac{2}{\frac{1}{4}}\right)=\pi -{\tan }^{-1}(8)$$