Question

sec 2x - tan 2x = tan(π/4-x)​

Answer 1

Given,

$sec2x-tan2x=tan\left(\frac{\pi }{4}-x\right)\\=\frac{1-2\cos \left(x\right)\sin \left(x\right)}{\cos \left(2x\right)}\\=\frac{\left(\cos \left(x\right)-\sin \left(x\right)\right)^2}{\cos ^2\left(x\right)-\sin ^2\left(x\right)}\\\mathrm{Factor}s:\\=\frac{\left(\cos \left(x\right)-\sin \left(x\right)\right)^2}{\left(\cos \left(x\right)+\sin \left(x\right)\right)\left(\cos \left(x\right)-\sin \left(x\right)\right)}$

$\mathrm{Manipulating\:right\:side}$

$\tan \left(\frac{\pi }{4}-x\right)\\\\=\frac{\cos \left(x\right)-\sin \left(x\right)}{\cos \left(x\right)+\sin \left(x\right)}\\=\frac{\cos ^2\left(x\right)+\sin ^2\left(x\right)-2\cos \left(x\right)\sin \left(x\right)}{\cos ^2\left(x\right)-\sin ^2\left(x\right)}\\=\frac{1-2\cos \left(x\right)\sin \left(x\right)}{\cos ^2\left(x\right)-\sin ^2\left(x\right)}$