Hi. Looking for help on another trig proof. Here's the problem.
Prove that:
sec^6x (secx.tanx) - sec^4x (secx.tanx) =sec^5x.tan^3x
This one's got me baffled. Thanks in Advance,
Mike Clemmons
What have you tried?
Can I suggest taking out $\displaystyle \sec x \tan x$ as a common factor first on the LHS
so
$\displaystyle (\sec x \tan x)(\sec^6 x- \sec^4 x)$
$\displaystyle (\sec x \tan x)(\sec^4 x(\sec^2 x- 1))$
$\displaystyle (\sec x \tan x)(\sec^4 x(\tan^2 x))$
The rest should be easy...