A) \[ta{{n}^{2}}\theta \]
B) \[co{{t}^{2}}\theta \]
C) \[se{{c}^{2}}\theta \]
D) \[cose{{c}^{2}}\theta \]
E) \[se{{c}^{3}}\theta \]
Correct Answer: C
Solution :
\[x={{\cos }^{3}}\theta ,y={{\sin }^{3}}\theta \] On differentiating w.r.t. \[\theta \]respectively \[\frac{dx}{d\theta }=-3{{\cos }^{2}}\theta \sin \theta \] and \[\frac{dy}{d\theta }=3{{\sin }^{2}}\theta \cos \theta \] Now, \[\frac{dy}{dx}=-\frac{3{{\sin }^{2}}\theta \cos \theta }{3{{\cos }^{2}}\theta \sin \theta }=-\tan \theta \] \[\therefore \] \[1+{{\left( \frac{dy}{dx} \right)}^{2}}=1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \]
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