NIMCET 2022 — Mathematics PYQ

To solve this, we convert the inverse trigonometric terms into a common form, typically ${\tan }^{-1}$.

1. Convert ${\csc }^{-1}\frac{5}{3}$ to ${\tan }^{-1}$

Let $\theta ={\csc }^{-1}\frac{5}{3}$, which implies $\csc \theta =\frac{5}{3}=\frac{\text{hypotenuse}}{\text{perpendicular}}$.

Using the Pythagorean theorem, the base is $\sqrt{5^2-3^2}=\sqrt{25-9}=\sqrt{16}=4$.

Therefore, $\tan \theta =\frac{\text{perpendicular}}{\text{base}}=\frac{3}{4}$.

So, ${\csc }^{-1}\frac{5}{3}={\tan }^{-1}\frac{3}{4}$.

2. Combine the terms

Now the expression becomes:

$$\cot \left({\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{2}{3}\right)$$

Using the identity ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$:

$${\tan }^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-(\frac{3}{4}\cdot \frac{2}{3})}\right)={\tan }^{-1}\left(\frac{\frac{9+8}{12}}{1-\frac{6}{12}}\right)$$

$$={\tan }^{-1}\left(\frac{\frac{17}{12}}{\frac{6}{12}}\right)={\tan }^{-1}\left(\frac{17}{6}\right)$$

3. Final Calculation

The expression is now:

$$\cot \left({\tan }^{-1}\frac{17}{6}\right)$$

Since $\cot ({\tan }^{-1}x)=\frac{1}{x}$:

$$\cot \left({\tan }^{-1}\frac{17}{6}\right)=\frac{1}{17/6}=\frac{6}{17}$$

Correct Option: A) $\frac{6}{17}$