MHT CET · Maths · Differentiation
If \(x=a \cos ^3 \theta \quad y=a \sin ^3 \theta\)
Then \(\sqrt{1+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2}=\)
- A \(\tan ^2 \theta\)
- B \(\sec ^2 \theta\)
- C \(\sec \theta\)
- D \(\tan \theta\)
Answer & Solution
Correct Answer
(C) \(\sec \theta\)
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{d}x}{\mathrm{d}\theta} = -3a \cos^2 \theta \sin \theta\) \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 3a \sin^2 \theta \cos \theta\)
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