Posted by **berhana** on Monday, November 6, 2006 at 10:05pm.

how do you start this equation i've been tryng it for 20min.

sec^6x(secxtanx)-sec^4x(secxtanx)=sec^5xtan^3x

ec^6x(secxtanx)-sec^4x(secxtanx)=sec^5xtan^3x

Factor out a sec^5 tan and divide thru. Left is

sec^2 x = Tan^2 x

Then this should reduce to

sin^2 x = cos^4 x

take the square root of each side

sin x= cos^2 x

change the cos squared to 1-sin^2

sin^2 x + sin x -1=0

Now you have a quadratic, use the quadratic equation.

## Answer This Question

## Related Questions

- Calc - Hello im trying to integrate tan^3 dx i have solved out the whole thing ...
- Calc - Hello im trying to integrate tan^3 dx i have solved out the whole thing ...
- Integration - Intergrate ¡ì sec^3(x) dx could anybody please check this answer. ...
- calculus - Did I do this problem right? Find the first and second derative-...
- Calculus 12th grade (double check my work please) - 2- given the curve is ...
- Math - Derivative of y=sec(tanx)? I don't get how it is secxsecx*secxtanx*...
- calculus (check my work please) - Not sure if it is right, I have check with the...
- calculus - find dy/dx y=ln (secx + tanx) Let u= secx + tan x dy/dx= 1/u * du/dx ...
- Calculus AP - I'm doing trigonometric integrals i wanted to know im doing step ...
- Trig - Find secx if sinx = -4/5 and 270 < x < 360. tan^2x+1=sec^2x (-4/5)^...

More Related Questions