((1+sin theta)^(2)+(1-sin theta)^(2))/(2cos20)=(1+sin theta theta)/(+sin2 theta)

If cos theta>sin theta>0 then I=int[ln[(1+sin2 theta)/(1-sin2 theta)]^(cos^(2)theta)+ln[(cos2 theta)/(1+sin2 theta)]]K sin2 theta ln[(cos theta+sin theta)/(cos theta-sin theta)]+m ln|cos2 theta|+c then k+m=

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

If (cos theta_(1))/(cos theta_(2))+(sin theta_(1))/(sin theta_(2))=(cos theta_(0))/(cos theta_(2))+(sin theta_(0))/(sin theta_(2))=1 , where theta_(1) and theta_(0) do not differ by can even multiple of pi , prove that (cos theta_(1)*cos theta_(0))/(cos^( 2)theta_(2))+(sin theta_(1)*sin theta_(0))/(sin^(2) theta_(2))=-1

If (cos theta_(1))/(cos theta_(2))+(sin theta_(1))/(sin theta_(2))=(cos theta_(0))/(cos theta_(2))+(sin theta_(0))/(sin theta_(2))=1 , where theta_(1) and theta_(0) do not differ by can even multiple of pi , prove that (cos theta_(1)*cos theta_(0))/(cos^( 2)theta_(2))+(sin theta_(1)*sin theta_(0))/(sin^(2) theta_(2))=-1

If (cos theta_(1))/(cos theta_(2))+(sin theta_(1))/(sin theta_(2))=(cos theta_(0))/(cos theta_(2))+(sin theta_(0))/(sin theta_(2))=1 , where theta_(1) and theta_(0) do not differ by can even multiple of pi , prove that (cos theta_(1)*cos theta_(0))/(cos^( 2)theta_(2))+(sin theta_(1)*sin theta_(0))/(sin^(2) theta_(2))=-1

The value of (2 (sin2 theta-2 cos^(2)theta-1))/(cos theta-sin theta-cos 3 theta +sin 3 theta)=

(1+sin2 theta-cos2 theta)/(1+sin2 theta+cos2 theta)=tan theta

The value of (2(sin2 theta+2cos^(2)theta-1))/(cos theta-sin theta-cos3 theta+sin3 theta)=

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

Prove that sin^(2)theta+sin^(2)2 theta+sin^(2)3 theta+....+sin^(2)n theta=(n)/(2)-(sin n theta cos(n+1)theta)/(2sin theta)