Question

If $ x={{\cos }^{3}}\theta $ and $ y={{\sin }^{3}}\theta , $ then $ 1+{{\left( \frac{dy}{dx} \right)}^{2}} $ is equal to:

• 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 $

Answer 1

Answer: (C)

$ 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 $