i.     \(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\theta \quad-\pi<\theta<\pi\)

\(\text {Range:}\ \ \tan ^{-1}(3 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right), \ \tan ^{-1}(10 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\)

 \(\Rightarrow \text { Both are monotonically increasing functions}\)

\(\Rightarrow\tan ^{-1}(3 x)+\tan ^{-1}(10 x) \text{ is also monotonically increasing with range }(-\pi, \pi)\) 

 \(\Rightarrow \text{ Only 1 solution exists (horizontal line will only cut graph once).}\)
 

ii.    \(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\dfrac{3 \pi}{4}\)

\(\tan \left(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)\right)=\tan \left(\dfrac{3 \pi}{4}\right)\)

\(\dfrac{\tan \left(\tan ^{-1}(3 x)\right)+\tan \left(\tan ^{-1}(10 x)\right)}{1-\tan \left(\tan ^{-1}(3 x)\right) \cdot \tan \left(\tan ^{-1}(10 x)\right)}=-1\)

 
\(x=\dfrac{1}{2}\ \ \text {or}\ \ -\dfrac{1}{15}\)

\(\text {Graph is monotonically increasing through } (0,0) \Rightarrow \ \Big(x \neq -\dfrac{1}{15} \Big)\)

\(\therefore x=\dfrac{1}{2}\)