NCERT Class 12 Maths · Chapter 5 Continuity And Differentiability · Exercise 5.6 Question 7

✅ Step-by-Step Solution

1. Differentiate both coordinates: dx/dt=\(\frac{\sin^{3}{\left(t \right)} \sin{\left(2 t \right)}}{\cos^{\frac{3}{2}}{\left(2 t \right)}} + \frac{3 \sin^{2}{\left(t \right)} \cos{\left(t \right)}}{\sqrt{\cos{\left(2 t \right)}}}\), dy/dt=\(- \frac{3 \sin{\left(t \right)} \cos^{2}{\left(t \right)}}{\sqrt{\cos{\left(2 t \right)}}} + \frac{\sin{\left(2 t \right)} \cos^{3}{\left(t \right)}}{\cos^{\frac{3}{2}}{\left(2 t \right)}}\).
2. Divide the two rates: dy/dx=(dy/dt)/(dx/dt)=\(\frac{- \frac{3 \sin{\left(t \right)} \cos^{2}{\left(t \right)}}{\sqrt{\cos{\left(2 t \right)}}} + \frac{\sin{\left(2 t \right)} \cos^{3}{\left(t \right)}}{\cos^{\frac{3}{2}}{\left(2 t \right)}}}{\frac{\sin^{3}{\left(t \right)} \sin{\left(2 t \right)}}{\cos^{\frac{3}{2}}{\left(2 t \right)}} + \frac{3 \sin^{2}{\left(t \right)} \cos{\left(t \right)}}{\sqrt{\cos{\left(2 t \right)}}}}\).
3. Simplify to \(- \frac{2 \cos{\left(2 t \right)} - 1}{\left(2 \cos{\left(2 t \right)} + 1\right) \tan{\left(t \right)}}\). This quotient requires dx/dt≠0; any cancelled zero requires a separate limiting or curve check.