`((cos(9 0^@)-theta)sin(9 0^@-theta))/(tan(9 0^@-theta)`

Prove that :(sin theta cos(90^(0)-theta)cos theta)/(sin(90^(0)-theta))+(cos theta sin(90^(0)-theta)sin theta)/(cos(90^(@)-theta))=1csc^(2)(90^(@)-theta)-tan^(2)theta=cos^(2)(90^(@)-theta)+cos^(2)theta

tantheta+tan(9 0^(@)-theta)=secthetaxxsec(9 0^(@)-theta)

Prove that :(sin theta)/(cos(3 theta))+(sin3 theta)/(cos9 theta)+(sin9 theta)/(cos27 theta)=(1)/(2)(tan27 theta-tan theta)

Prove that : sintheta/cos(3theta)+sin(3theta)/cos(9theta)+sin(9theta)/cos(27theta)=1/2(tan(27theta)-tan(theta))

(sin theta)/(cos(3theta))+(sin(3theta))/(cos(9theta))+(sin(9theta))/(cos(27theta))+(sin(27theta))/(cos(81theta))=

(sin theta)/(cos(3theta))+(sin(3theta))/(cos(9theta))+(sin(9theta))/(cos(27theta))+(sin(27theta))/(cos(81theta))=

If k_(1)=tan27 theta-tan theta and k_(2)=(sin theta)/(cos3 theta)+(sin3 theta)/(cos9 theta)+(sin9 theta)/(cos27 theta)

(sin theta)/(cos3 theta)+(sin3 theta)/(cos9 theta)+(sin9 theta)/(cos27 theta)=lambda[(Tan mu theta)-Tan theta]" then "mu-50 lambda =

(sin theta)/(cos3 theta)+(sin3 theta)/(cos9 theta)+(sin9 theta)/(cos27 theta)+(sin27 theta)/(cos81 theta)=

Integration using rigonometric identities : int(sin theta)/(cos3 theta)+(sin 3 theta)/(cos9 theta)+( sin9 theta)/(cos 27theta)d theta=...+c