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[" 24."(sin3 theta-cos3 theta)/(sin thet...

[" 24."(sin3 theta-cos3 theta)/(sin theta+cos theta)+1=],[[" (A) "2sin2 theta," (B) "2cos2 theta," (C) "tan2 theta," (D) "cot29]]

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(3sin theta-sin3 theta)/(3cos theta+cos3 theta)=

(3cos theta+cos3 theta)/(3sin theta-sin3 theta)=

(sin 3theta)/(sin theta)-(cos 3theta)/(cos theta)=

(sin theta-cos theta+1)/(sin theta+cos theta-1)=(1+sin theta)/(cos theta)

Prove that: (sin^(3) theta + cos^(3) theta)/(sin theta + cos theta) = 1-sin theta cos theta

(sin3 theta-sin theta)/(cos theta-cos3 theta)=cot2 theta

(sin theta+cos theta)(1-sin theta cos theta)=sin^3 theta+cos^3 theta

Simplify: (sin^(3) theta + cos^(3) theta)/(sin theta + cos theta) + sin theta cos theta

(cos theta+cos3 theta)/(sin theta-sin3 theta) is equal to:

( cos 3 theta - sin 3 theta )/( cos theta+ sin theta ) =