Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

We use the identity ${\mathrm{Tan}}^{-1}(x)-{\mathrm{Tan}}^{-1}(y)={\mathrm{Tan}}^{-1}(\frac{x-y}{1+xy})$.

${\mathrm{Tan}}^{-1}\left(\frac{1}{2K^2}\right)={\mathrm{Tan}}^{-1}\left(\frac{(2K+1)-(2K-1)}{1+(2K+1)(2K-1)}\right)={\mathrm{Tan}}^{-1}(2K+1)-{\mathrm{Tan}}^{-1}(2K-1)$.

The sum telescopes to ${\lim }_{K\to \infty }{\mathrm{Tan}}^{-1}(2K+1)-{\mathrm{Tan}}^{-1}(1)=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$.

Therefore, $\theta =\frac{\pi }{4}$, and $\mathrm{Tan}\theta =\mathrm{Tan}\frac{\pi }{4}=1$.