If $x=\sec heta-\cos heta, y=\sec ^n heta-\cos ^n heta$ then $\frac{d y}{d x}=$

Question

If $x=\sec 	heta-\cos 	heta, y=\sec ^n 	heta-\cos ^n 	heta$ then $\frac{d y}{d x}=$

Aicharya · Accepted Answer

(b) We have, $ x=\sec 	heta-\cos 	heta $ differential w.r.t $	heta$ $ \begin{aligned} \frac{d x}{d 	heta} & =\sec 	heta 	an 	heta+\sin 	heta \ & =\sec 	heta \cdot \frac{\sin 	heta}{\cos 	heta}+\sin 	heta \ & =	an 	heta(\sec 	heta+\cos 	heta) \end{aligned} $ Now, $\quad y=\sec ^n 	heta-\cos ^n 	heta$ differential w.r.t ' $	heta$ ' $ \begin{aligned} \frac{d y}{d 	heta} & =n \sec ^{n-1} 	heta \cdot \sec 	heta 	an 	heta-n \cos ^{n-1} 	heta(-\sin 	heta) \ & =n \sec ^n 	heta \frac{\sin 	heta}{\cos 	heta}+n \cos ^{n-1} 	heta \cdot \sin 	heta \ & =n \frac{\sin 	heta}{\cos 	heta}\left(\sec ^n 	heta+\cos ^n 	hetaight) \ & =n 	an 	heta\left(\sec ^n 	heta+\cos ^n 	hetaight) \end{aligned} $ Now, $ \begin{aligned} & \frac{d y}{d x}=\frac{n 	an 	heta\left(\sec ^n 	heta+\cos ^n 	hetaight)}{	an 	heta(\sec 	heta+\cos 	heta)} \ & \frac{d y}{d x}=\frac{n\left(\sec ^n 	heta+\cos ^n 	hetaight)}{(\sec 	heta+\cos 	heta)} \end{aligned} $ Squaring both side $\begin{aligned} & \left(\frac{d y}{d x}ight)^2=\frac{n^2\left(\sec ^n 	heta+\cos ^n 	hetaight)^2}{(\sec 	heta+\cos 	heta)^2} \ & \left(\frac{d y}{d x}ight)^2=\frac{n^2\left(\sec ^{2 n} 	heta+\cos ^{2 n} 	heta+2ight)}{\left(\sec ^2 	heta+\cos ^2 	heta+2ight)} \ & =\frac{n^2\left(\sec ^{2 n} 	heta+\cos ^{2 n} 	heta-2+4ight)}{\left(\sec ^2 	heta+\cos ^2 	heta-2+4ight)} \ & =\frac{n^2\left[\left(\sec ^n 	heta-\cos ^n 	hetaight)^2+4ight]}{(\sec 	heta-\cos 	heta)^2+4} \ & \left(\frac{d y}{d x}ight)^2=\frac{n^2\left(y^2+4ight)}{\left(x^2+4ight)} \ & \Rightarrow \frac{d y}{d x}=n \sqrt{\frac{y^2+4}{x^2+4}} \ & \end{aligned}$

If $x=\sec \theta-\cos \theta, y=\sec ^n \theta-\cos ^n \theta$ then $\frac{d y}{d x}=$

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