यदि `A=[(cosx,-sinx),(sinx,cosx)]` तो दिखाइए कि `(A^(-1))^(-1)=A`

|[cosx,-sinx],[sinx,cosx]|

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

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

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

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

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

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

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