NIMCET 2022 — Mathematics PYQ

Solution 
Step 1: Convert ${\text{cosec}}^{-1}\frac{5}{3}$ to ${\text{tan}}^{-1}$ 
Let $\theta ={\text{cosec}}^{-1}\frac{5}{3}$. This implies that $\text{cosec}\theta =\frac{5}{3}$. From the definition of cosecant, $\text{sin}\theta =\frac{1}{\text{cosec}\theta }=\frac{3}{5}$. Using the Pythagorean identity, ${\text{cos}}^2\theta +{\text{sin}}^2\theta =1$, it is found that ${\text{cos}}^2\theta =1-{\left(\frac{3}{5}\right)}^2=1-\frac{9}{25}=\frac{16}{25}$. Therefore, $\text{cos}\theta =\frac{4}{5}$ (assuming $\theta$ is in the first quadrant, where ${\text{cosec}}^{-1}$ is defined). Then, $\text{tan}\theta =\frac{\text{sin}\theta }{\text{cos}\theta }=\frac{3/5}{4/5}=\frac{3}{4}$. So, ${\text{cosec}}^{-1}\frac{5}{3}={\text{tan}}^{-1}\frac{3}{4}$. 
Step 2: Apply the ${\text{tan}}^{-1}$ addition formula 
The expression becomes $\text{cot}\left({\text{tan}}^{-1}\frac{3}{4}+{\text{tan}}^{-1}\frac{2}{3}\right)$. Using the formula ${\text{tan}}^{-1}x+{\text{tan}}^{-1}y={\text{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$, it is found that: ${\text{tan}}^{-1}\frac{3}{4}+{\text{tan}}^{-1}\frac{2}{3}={\text{tan}}^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\cdot \frac{2}{3}}\right)$. The numerator is $\frac{3}{4}+\frac{2}{3}=\frac{9+8}{12}=\frac{17}{12}$. The denominator is $1-\frac{6}{12}=1-\frac{1}{2}=\frac{1}{2}$. So, ${\text{tan}}^{-1}\frac{3}{4}+{\text{tan}}^{-1}\frac{2}{3}={\text{tan}}^{-1}\left(\frac{17/12}{1/2}\right)={\text{tan}}^{-1}\left(\frac{17}{12}\cdot 2\right)={\text{tan}}^{-1}\frac{17}{6}$. 
Step 3: Calculate the cotangent 
The expression is now $\text{cot}\left({\text{tan}}^{-1}\frac{17}{6}\right)$. Since $\text{cot}({\text{tan}}^{-1}x)=\frac{1}{x}$, it is found that: $\text{cot}\left({\text{tan}}^{-1}\frac{17}{6}\right)=\frac{1}{17/6}=\frac{6}{17}$. 
Final Answer 
The final answer is $\boxed{\text{6/17}}$.