[3cos theta=5sin alpha" fund "],[(5sin theta-2sec^(3)theta+2cos theta)/(5sin theta+2sec^(3)theta-2cos theta)]

If 3cos theta=5sin theta, then the value of (5sin theta-2sec^(3)theta+2cos theta)/(5sin theta+2sec^(3)theta-2cos theta) is (271)/(979) (b) (316)/(2937) (c) (542)/(2937) (d) None of these

If 3cos theta=5sin theta, then find the value of 5sin theta-2sec^(3)theta+2cos theta5sin theta+2sec^(3)theta-2cos theta

(cos theta-sin theta) / (cos theta + sin theta) = sec2 theta-tan2 theta

4 sin 5 theta cos 3 theta sin 2 theta

If tan theta=1 then (8sin theta +5 cos theta)/(sin^3theta-2cos^3theta+7cos theta)=

cos theta-cos3 theta+cos5 theta-cos7 theta=4sin theta sin4 theta cos2 theta

(sin5 theta-2sin3 theta + sin theta) / (cos5 theta-cos theta) = tan theta

2cos theta-cos3 theta-cos5 theta-16cos^(3)theta sin^(2)theta=

2cos theta-cos3 theta-cos5 theta-16cos^(3)theta sin^(2)theta=

(sin theta+sin3 theta+sin5 theta)/(cos theta+cos3 theta+cos5 theta)=tan3 theta