If `(1-t a nthetat a ntheta1)(1t a ntheta-t a ntheta1)=[a-bb a]` , then `a=cos2theta` b. `a=1` c. `b=s in2theta` d. `b=-1`

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,-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

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

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

Evaluate: intsqrt(t a ntheta)\ d theta

Evaluate: intsqrt(t a ntheta)\ d theta

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

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

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

If a=s e ctheta-t a ntheta and b=cos e ctheta+cottheta , then show that a b+a-b+1=0.