If` cos theta_(1)+cos theta_(2)+cos theta_(3)+cos theta_(4)+cos theta_(5)=5 then , sin theta_(1)+sin theta_(2)+sin theta_(3)+sin theta_(4)+sin theta_(5)=`

If cos theta_(1)+cos theta_(2)+cos theta_(3)=3 then sin^(2)theta_(1)+sin^(4)theta_(2)+sin^(6)theta_(3)=

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)+cos theta_(3)=-3 then the value of sin((theta_(1))/(2))+sin((theta_(2))/(2))+sin((theta_(3))/(2))=

If sin theta_(1)+sin theta_(2)+sin theta_(3), then find the value of cos theta_(1)+cos theta_(2)+cos theta_(3)

cos (2 theta) cos ((theta) / (2)) - cos (3 theta) cos ((9 theta) / (2)) = sin (5 theta) sin ((5 theta) / (2))

If hat harr ABC having vertices A(a cos theta_(1),a sin theta_(1)),B(a cos theta_(2)a sin theta_(2)), and C(a cos theta_(3),a sin theta_(3)) is equilateral,then prove that cos theta_(1)+cos theta_(2)+cos theta_(3)=sin theta_(1)+sin theta_(2)+sin theta_(3)=0

If theta_(1),theta_(2),theta_(3)....,theta_(n) are in A.P. then (sin theta_(1)+sin theta_(2)+...+sin theta_(n))/(cos theta_(1)+cos theta_(2)+...+cos theta_(n))=

" 4."(sin theta)/(1+cos theta)+(1+cos theta)/(sin theta)=(2)/(sin theta)