NIMCET 2022 — Mathematics PYQ

1. The given equation is ${\cos }^{-1}\frac{x}{2}+{\cos }^{-1}\frac{y}{3}=\varphi$​. 
2. The formula for the sum of inverse cosines is ${\cos }^{-1}A+{\cos }^{-1}B={\cos }^{-1}\left(AB-\sqrt{1-A^2}\sqrt{1-B^2}\right)$​. 
3. Applying the formula, it is obtained that ${\cos }^{-1}\left(\frac{x}{2}\cdot \frac{y}{3}-\sqrt{1-{\left(\frac{x}{2}\right)}^2}\sqrt{1-{\left(\frac{y}{3}\right)}^2}\right)=\varphi$​. 
4. Taking the cosine of both sides, it is found that $\frac{xy}{6}-\sqrt{1-\frac{x^2}{4}}\sqrt{1-\frac{y^2}{9}}=\cos \varphi$​. 
5. Rearranging the terms, it is obtained that $\frac{xy}{6}-\cos \varphi =\sqrt{1-\frac{x^2}{4}}\sqrt{1-\frac{y^2}{9}}$​. 
6. Squaring both sides, it is found that ${\left(\frac{xy}{6}-\cos \varphi \right)}^2=\left(1-\frac{x^2}{4}\right)\left(1-\frac{y^2}{9}\right)$​. 
7. Expanding both sides, it is obtained that $\frac{x^2{y}^2}{36}-2\cdot \frac{xy}{6}\cos \varphi +{\cos }^2\varphi =1-\frac{y^2}{9}-\frac{x^2}{4}+\frac{x^2{y}^2}{36}$​. 
8. Simplifying the equation, it is found that $\frac{x^2{y}^2}{36}-\frac{xy}{3}\cos \varphi +{\cos }^2\varphi =1-\frac{y^2}{9}-\frac{x^2}{4}+\frac{x^2{y}^2}{36}$​. 
9. Canceling the common term $\frac{x^2{y}^2}{36}$​ from both sides, it is obtained that $-\frac{xy}{3}\cos \varphi +{\cos }^2\varphi =1-\frac{y^2}{9}-\frac{x^2}{4}$​. 
10. Multiplying the entire equation by $36$​ to eliminate fractions, it is found that $-12xy\cos \varphi +36{\cos }^2\varphi =36-4y^2-9x^2$​. 
11. Rearranging the terms to match the desired expression, it is obtained that $9x^2-12xy\cos \varphi +4y^2=36-36{\cos }^2\varphi$​. 
12. Factoring out $36$​ and using the identity $1-{\cos }^2\varphi ={\sin }^2\varphi$​, it is found that $9x^2-12xy\cos \varphi +4y^2=36(1-{\cos }^2\varphi )=36{\sin }^2\varphi$​. 

Final Answer 
The expression $9x^2-12xy\cos \varphi +4y^2$​ is equal to $36{\sin }^2\varphi$​.