If `f(theta)=|[cos^(2)theta,cos theta sin theta,-sin theta],[cos theta sin theta,sin^(2)theta,cos theta],[sin theta,-cos theta,0]|`, then `f((pi)/(12))`

If f(theta)=|(cos^(2)theta ,cos theta sin theta, -sin theta),(cos theta sin theta, sin^(2)theta,cos theta),(sin theta,-cos theta,0)| then, f((pi)/(6))+f((pi)/(3))+f((pi)/(2))+f((2pi)/(3))+f((5pi)/(6))+f(pi)+……+f((53pi)/(6)) is equal to

If f (theta) = [[cos^(2) theta , cos theta sin theta,-sin theta],[cos theta sin theta , sin^(2) theta , cos theta ],[sin theta ,-cos theta , 0]] "then " f (pi/7) is

cos theta+sin theta=cos2 theta+sin2 theta

[[ Simplify :cos thetacos theta,sin theta-sin theta,cos theta]]+sin theta[[sin theta,-cos thetacos theta,sin theta]]

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

Simplify: cos theta[cos theta sin theta-sin theta cos theta]+sin theta[sin theta-cos theta cos theta sin theta]

Simplify cos theta[(cos theta,sin theta),(-sin theta,cos theta)]+sin theta[(sin theta,-cos theta),(cos theta,sin theta)]

If cos theta+sin theta=sqrt(2), then cos theta-sin theta=

Consider f(theta)=|[tan theta,cos theta,sin theta],[cos theta,sin theta,tan theta],[sin theta,tan theta,cos theta]| then f((pi)/(4))+f(0) = ?