`A=|{:(cosx,-sinx),(sinx,cosx):}|`, then show that `(A^(-1))^(-1)=A.`

If A=[{:(cosx,sinx),(-sinx,cosx)] then find |A|

If A=[{:(cosx,sinx),(-sinx,cosx)] then find |A|

Expand | (cosx, -sinx),(sinx , cosx)|

cosx{(cosx)/(1-sinx)+(1-sinx)/(cosx)}, is

cosx{(cosx)/(1-sinx)+(1-sinx)/(cosx)}, is

((sinx - cosx)/(sinx + cosx))

If y=(1+sinx-cosx)/(1+sinx+cosx), " show that, " (dy)/(dx)=(1)/(1+cos x).

If F(x)=[(cosx,-sinx,0),(sinx,cosx,0),(0,0,1)] and G(x)=[(cosx,0,sinx),(0,1,0),(-sinx,0,cosx)] then [F(x)G(x)]^(-1) =

int(1+sinx)/(sinx(1+cosx))dx