`cos thetaxxtan theta=sin theta`

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

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)/(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 / 7) is

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

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

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

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

(1-cos theta)/(sin theta)=(sin theta)/(1+cos theta)

(1-cos theta)/(sin theta)=(sin theta)/(1+cos theta)