A) \[\frac{{{n}^{2}}({{y}^{2}}+4)}{{{x}^{2}}+4}\]
B) \[\frac{{{n}^{2}}({{y}^{2}}-4)}{{{x}^{2}}}\]
C) \[n\frac{({{y}^{2}}-4)}{{{x}^{2}}-4}\]
D) \[{{\left( \frac{ny}{x} \right)}^{2}}-4\]
Correct Answer: A
Solution :
Differentiate given equations w.r.t. \[\theta ,\]we get \[\frac{dy}{d\theta }=n{{\sec }^{n-1}}\theta .\sec \theta .\tan \theta \] \[-n.{{\cos }^{n-1}}\theta (-sin\theta )\] \[=\tan \theta (se{{c}^{2}}\theta +co{{s}^{n}}\theta )\] \[\frac{dx}{d\theta }=\sec \theta \tan \theta +\sin \theta =\tan \theta (sec\theta +cos\theta )\] \[\therefore \] \[\frac{dy}{dx}=\frac{n\tan \theta (se{{c}^{n}}\theta +{{\cos }^{n}}\theta )}{\tan \theta (sec\theta +cos\theta )}\] \[=\frac{n(se{{c}^{n}}\theta +{{\cos }^{n}}\theta )}{(sec\theta +cos\theta )}\] \[\therefore \] \[{{\left( \frac{dy}{dx} \right)}^{2}}=\frac{{{n}^{2}}{{(se{{c}^{n}}\theta +co{{s}^{n}}\theta )}^{2}}}{{{(sec\theta +cos\theta )}^{2}}}\] \[=\frac{{{n}^{2}}\{{{(se{{c}^{n}}\theta -co{{s}^{n}}\theta )}^{2}}+4\}}{{{(sec\theta -\cos \theta )}^{2}}+4}\] \[=\frac{{{n}^{2}}({{y}^{2}}+4)}{({{x}^{2}}+4)}\]
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