Hi, I'm obviously missing something here as I am only getting further awway from the answer.
show that d/dx (cos x + sin x/ cos x - sin x) = sec^2(pi/4 +x)
Thanks in advance
let's do something with the secant mess first ...
$\displaystyle \sec^2\left(\frac{\pi}{4}+x\right) =$
$\displaystyle 1 + \tan^2\left(\frac{\pi}{4}+x\right) =$
$\displaystyle 1 + \left[ \frac{\tan\left(\frac{\pi}{4}\right) + \tan{x}}{1 - \tan\left(\frac{\pi}{4}\right)\tan{x}}\right]^2 =$
$\displaystyle 1 + \left[\frac{1 + \tan{x}}{1 - \tan{x}}\right]^2 =$
$\displaystyle 1 + \left[\frac{\cos{x}+\sin{x}}{\cos{x}-\sin{x}}\right]^2 =$
$\displaystyle \left[\frac{\cos{x}-\sin{x}}{\cos{x}-\sin{x}}\right]^2+ \left[\frac{\cos{x}+\sin{x}}{\cos{x}-\sin{x}}\right]^2 =$
$\displaystyle \frac{(\cos{x}-\sin{x})^2 + (\cos{x}+\sin{x})^2}{(\cos{x}-\sin{x})^2}$
now take the derivative of the left side and see if it matches up.