Question

differentiate 1-cos2x/1+cos2x (pls write the steps)

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: find : \\ \frac{d}{dx} ( \frac{1 - cos2x}{1 + cos2x} ) \\ \\ we \: know \: that \\ cos2x = cos {}^{2} x - sin {}^{2} x \\ \\ y = \frac{1 - cos2x}{1 + cos2x} \\ \\ = \frac{1 - (cos {}^{2}x - sin {}^{2}x) }{1 + cos {}^{2} - sin {}^{2}x } \\ \\ = \frac{1 - cos {}^{2} x + sin {}^{2}x }{1 - sin {}^{2} x + cos {}^{2} x} \\ \\ = \frac{sin {}^{2} x + sin {}^{2} x}{cos {}^{2}x + cos {}^{2} x} \\ \\ = \frac{sin {}^{2} x}{cos {}^{2}x } \\ \\ y= tan {}^{2} x \\ \\ differenntiating \: both \: sides \\ we \: get \\ \frac{dy}{dx} = \frac{d}{dx} (tan {}^{2} x) \\ \\ = \frac{d}{dx} (tan \: x.tan \: x) \\ \\ we \: know \: that \\ \frac{du}{dv} = u. \frac{dv}{dx} + v. \frac{du}{dx} \\ \\ = tan \: x. \frac{d}{dx} (tan \: x) + tan \: x. \frac{d}{dx} (tan \: x) \\ \\ we \: know \: that \\ \frac{d}{dx} (tan \: x) = sec {}^{2} x \\ \\ = tan \: x.sec {}^{2} x + tan \: x.sec {}^{2} x \\ \\ = 2.tan \: x.sec {}^{2} x \\ or \\ = 2. \frac{sin \: x}{cos {}^{3}x }$