Simplify `costheta[[costheta,sintheta],[-sintheta,costheta]+sintheta[[sintheta,-costheta],[costheta,sintheta]]`

Show that AB=BA if A=[[costheta,sintheta],[-sintheta,costheta]],B=[[costheta,-sintheta],[sintheta,costheta]]

Show that AB=BA where, A=[[costheta,sintheta],[-sintheta,costheta]],B=[[costheta,-sintheta],[sintheta,costheta]]

Show that: |(sintheta,costheta),(-costheta,sintheta)| =1

Answer the following questions.If A=[[costheta,sintheta],[-sintheta,costheta]] ,prove that A^n=[[cos ntheta,sin ntheta],[-sin ntheta,cos ntheta]] , for all n in N

(sintheta+sin 2theta)/(1+costheta+cos2theta)=

sintheta/(1-cottheta)+costheta/(1-tantheta)=

inte^(sintheta)[log(sintheta)+cosec^(2)theta]costheta"d"theta=

If costheta-sintheta=sqrt2sintheta then costheta+sintheta is equal to

int((sintheta+costheta))/sqrt(sin2theta)"d"theta=

Solve the following: Fing theta ,if [[sintheta],[costheta],[theta]][sintheta,costheta,theta]=[17]