this problem has been bugging me for a long time
1/1-secx + 1/1+secx
i keep ending up with 2/1-sec^2(x)
but the answer is -2cot^2(x)
can anybody help what am i doing wrong
Did you know: $\displaystyle 1 + tan^2 x = sec^2 x $?
If you don't want to use that identity I suggest you write the expression as $\displaystyle \frac{1}{1-\sec x} + \frac{1}{1+\sec x} = \frac{1}{\frac{\cos x -1}{\cos x}} + \frac{1}{\frac{\cos x +1}{\cos x}} = \frac{\cos x}{\cos x -1}+ \frac{\cos x}{\cos x +1}$ first.
so, you have $\displaystyle \frac{2}{1-sec^{2}x}$
$\displaystyle sex^2x-tan^2x=1$
so,
$\displaystyle tan^2x = sec^2x -1$
$\displaystyle -tan^2x = -sec^2x+1 =1-sec^2x$
so your denominator is equivalent to $\displaystyle -tan^2x$, thus giving you
$\displaystyle \frac{2}{-tan^2x} = -2cot^2x$ [because $\displaystyle cotx = \frac{1}{tanx}$]