1/(secx + tanx) + 1/(secx-tanx) = 2secx

[tdotgirl]

1 / (sec x + tan x) + 1 / (sec x - tan x) = 2 sec x

1 / ((1/cos x) + (sin x/cos x)) + 1 / ((1/cos x) - (sin x / cos x)) = 2/cosx

cosx + cosx/sinx + cosx - cosx/sinx = 2/cosx

2cosx = 2/cosx

i don't know where to go from here. did i make a mistake somewhere?

 

[tkhunny]

Switching to sine and cosine is NOT the ultimate solution. There are many other avenues of attack. In this case, why not just find a common denominator and add on the left-hand side?

\(\displaystyle \frac{[sec(x)-tan(x)]+[sec(x)+tan(x)]}{sec^{2}(x)-tan^{2}(x)}\)

Simplify that and see what drops out.

Rule of Thumb: IF ALL ELSE FAILS, switch to sine and cosine.

 

[tdotgirl]

thanks so much for the help and useful tips!!