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

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- Apr 25th 2010, 04:40 PMsomething3kSimplifying math identity
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 - Apr 25th 2010, 05:17 PMGusbob
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. - Apr 25th 2010, 05:57 PMharish21
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 $\displaystyle cotx = \frac{1}{tanx}$]