Prove this

[Drizzt]

i need to prove this trig equation:

1/1-secx + 1/1+secx = -2cot^2x

yes, the cot is squared. any help is appreciated.

 

[stapel]

Assuming you mean:

. . . . .$\displaystyle \large{\frac{1}{1\,-\,\sec{(x)}}\;+\;\frac{1}{1\,+\,\sec{(x)}}\;=\;-2\cot^2{(x)}}$

...a good first step might be to convert to a common denominator on the left, and combine the fractions. See where this takes you.

Eliz.

 

[galactus]

There are always different approaches at trig identities, but you could try

$\displaystyle \frac{1}{1-sec(x)}+\frac{1}{1+sec(x)}$

$\displaystyle \frac{1+sec(x)+1-sec(x)}{(1-sec(x))(1+sec(x))}$

$\displaystyle \frac{2}{(1-sec(x))(1+sec(x))}$

$\displaystyle \frac{2}{1-sec^{2}(x)}$

$\displaystyle \frac{-2}{sec^{2}(x)-1}$

Since $\displaystyle sec^{2}(x)-1=tan^{2}(x)$, we have

$\displaystyle \frac{-2}{tan^{2}(x)}=-2cot^{2}(x)$

 

[Drizzt]

yup, thats what i got now. thanks you guys.