`(2 sin theta*cos theta - cos theta)/(1-sin theta+sin^2 theta-cos^2 theta) = cot theta`

(2sin theta*cos theta-cos theta)/(1-sin theta+sin^(2)theta-cos^(2)theta)=cot theta

(sin3 theta+cos3 theta)/(cos theta-sin theta)=1+2xx sin2 theta

Evaluate the following determinants: (b) |(cos theta, -sin theta),(sin theta, cos theta)| = cos theta (cos theta) - sin theta(-sin theta) = cos^(2) theta + sin^(2) theta = 1

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

2(cos theta+cos2 theta)+(1+2cos theta)sin2 theta=2sin theta

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

The value of (cos^6 theta+sin^6 theta+ 3sin^2 theta cos^2 theta)/(cosec theta sec theta(sin theta+cos theta-1)(sin theta+cos theta+1)) is : (cos^6 theta+sin^6 theta+ 3sin^2 theta cos^2 theta)/(cosec theta sec theta(sin theta+cos theta-1)(sin theta+cos theta+1)) का मान ज्ञात कीजिए?

1-(sin^(2) theta)/(1+cos theta) + (1+cos theta)/(sin theta) - (sin theta)/(1-cos theta) - (1)/(sec theta)=

(sin theta)/(cot theta+cos ec theta)-(sin theta)/(cot theta-cos ec theta)=