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

Prove that, (sin 3 theta)/(sin theta)-(cos 3 theta)/(cos theta)=2

(sin 3 theta)/(sin theta)-(cos 3 theta)/(cos theta)= A) sin 2 theta B) cos 4 theta C)1D)2

Solve (3 sin theta-sin 3 theta)/(sin theta)+(cos 3 theta)/(cos theta)=1 .

Solve (3 sin theta-sin 3 theta)/(sin theta)+(cos 3 theta)/(cos theta)=1 .

Solve (3 sin theta-sin 3 theta)/(sin theta)+(cos 3 theta)/(cos theta)=1 .

Prove that (sin 3 theta)/( sin theta) - (cos 3 theta)/(cos theta) = 2

The value of the expression ((sin 3theta)/sin theta)^2-((cos3theta)/cos theta)^2=

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

If A=6 sin20^(@) - 8 sin^(3) 20^@ , B= 8 cos^(3) 20^(@)-6 cos 20^(@) and C= ( sin 3 theta )/( sin theta ) - ( cos 3 theta )/( cos theta ) then

((sin3 theta)/(sin theta))^(2)-((cos3 theta)/(cos theta))^(2)=a cos2 theta wherever it id defined.Then the value of a is