Need the fundamental identities here.
tan(X) = sin(X)/cos(X)
sec(X) = 1/cos(X)
so
tan(X)/sec(X)
same as,
sin(X)/cos(X) * cos(X)/1
cancel the cos(X)
= sin(X)
---------------simplest form
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XtanX dx formula uv - int v du u = x du = dx dv = tanX dx v = ln(secX) x ln(secX) - int ln(secx) dx = X ln(secx) - x ln(secx) - x + C -----------------------------------------
to simplify Cosx=Sinx Tanx you should remember your fundamental and pythagorean identities.. Cosx + Sinx Tanx Cosx + Sinx (Sinx/Cosx) <---------- From Tanx= Sinx/Cosx Cosx + Sin2x/ Cos x <------------- do the LCD Cosx (Cosx/Cosx) + Sin2x/Cosx (Cos2x+Sin2x)/Cosx 1/Cosx <--------- From Sin2x + Cos2x =1 or Secx <-------- answer Comment if you have questions...:))
d/dx(uv)=u*dv/dx+v*du/dxd/dx(secxtanx)=secx*[d/dx(tanx)]+tanx*[d/dx(secx)]-The derivative of tanx is:d/dx(tan u)=[sec(u)]2*d/dx(u)d/dx(tan x)=[sec(x)]2*d/dx(x)d/dx(tan x)=[sec(x)]2*(1)d/dx(tan x)=(sec(x))2=sec2(x)-The derivative of secx is:d/dx(sec u)=[sec(u)tan(u)]*d/dx(u)d/dx(sec x)=[sec(x)tan(x)]*d/dx(x)d/dx(sec x)=[sec(x)tan(x)]*(1)d/dx(sec x)=sec(x)tan(x)d/dx(secxtanx)=secx*[sec2(x)]+tanx*[sec(x)tan(x)]d/dx(secxtanx)=sec3(x)+sec(x)tan2(x)
You can take out any constant from a derivative. In other words, this is the same as 5 times the derivative of sec x.
It is minus 1 I did this: sinx/cos x = tan x sinx x = cosx tanx you have (x - sinxcosx) / (tanx -x) (x- cos^2 x tan x)/(tanx -x) let x =0 -cos^2 x (tanx) /tanx = -cos^x -cos^2 (0) = -1