If `|(1,-t a ntheta),(t a ntheta,1)| |(1,t a ntheta),(-t a ntheta,1)|^-1=|[a,-b],[b ,a]|` , then a. `a=cos2theta` b. `a=1` c. `b=sin2theta` d. `b=-1`

If [(1,-tantheta),(tantheta,1)][(1,tantheta),(-tantheta,1)]^(-1)=[(a,-b),(b, a)] , then a=1,\ \ b=0 (b) a=cos2\ theta,\ \ b=sin2\ theta (c) a=sin2\ theta,\ \ b=cos2theta (d) none of these

Solve the equation : (1-t a ntheta)(1+sin2theta)=1+t a ntheta

Evaluate: intsqrt(t a ntheta)\ d theta

Prove that: tan4theta=(4t a ntheta(1-tan^2theta))/(1-6tan^2theta+tan^4theta)

Prove that: (sin2theta)/(1+cos2theta)=t a ntheta

Prove that: (sin2theta)/(1+cos2theta)=t a ntheta

Solve the following equation: t a ntheta+t a n2theta+t a n3theta=0

Solve: sqrt(2) sec theta+t a ntheta=1

Evaluate: int{sqrt(t a ntheta)+sqrt(cottheta)}\ dtheta

Prove the following identity: costheta(t a ntheta+2)(2t a ntheta+1)=2s e ctheta+5sintheta