Question
Diffrentiate the following w.r.t.x
$\frac{1+\sin x^{\circ}}{1-\sin x^{\circ}}$
$\frac{1+\sin x^{\circ}}{1-\sin x^{\circ}}$
Differentiating w.r.t. $x$, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left[\frac{1+\sin \left(\frac{\pi x}{180}\right)}{1-\sin \left(\frac{\pi x}{180}\right)}\right) \\ & '=\frac{\left[1-\sin \left(\frac{\pi x}{180}\right)\right] \cdot \frac{d}{d x}\left[1+\sin \left(\frac{\pi x}{180}\right)\right]-\left[1+\sin \left(\frac{\pi x}{180}\right)\right] \cdot \frac{d}{d x}\left[1-\sin \left(\frac{\pi x}{180}\right)\right]}{\left[1-\sin \left(\frac{\pi x}{180}\right)\right]^2} \\ & =\frac{\left[1-\sin \left(\frac{\pi x}{180}\right)\right] \cdot\left[0+\cos \left(\frac{\pi x}{180}\right) \cdot \frac{d}{d x}\left(\frac{\pi x}{180}\right)\right]-\left[1+\sin \left(\frac{\pi x}{180}\right)\right] \cdot\left[0-\cos \left(\frac{\pi x}{180}\right) \cdot \frac{d}{d x}\left(\frac{\pi x}{180}\right)\right]}{\left[1-\sin \left(\frac{\pi x}{180}\right)\right]^2} \\ & =\frac{\left(1-\sin x^{\circ}\right)\left[\left(\cos x^{\circ}\right) \times \frac{\pi}{180} \times 1\right]-\left(1+\sin x^{\circ}\right)\left[\left(-\cos x^{\circ}\right) \times \frac{\pi}{180} \times 1\right]}{\left(1-\sin x^{\circ}\right)^2} \\ & =\frac{\frac{\pi}{180} \cos x^{\circ}\left(1-\sin x^{\circ}+1+\sin x^{\circ}\right)}{\left(1-\sin x^{\circ}\right)^2} \\ & =\frac{\pi \cos x^{\circ}}{90\left(1-\sin x^{\circ}\right)^2} . \\ & \end{aligned}$
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