`sin theta+cos theta=1` Find `(sin theta-cos theta)^(2)`

If sin theta+cos theta=1 then sin theta cos theta=

Prove each of the following identities : (i) (sin theta - cos theta)/(sin theta + cos theta) + ( sin theta+ cos theta)/(sin theta - cos theta) = (2)/((2 sin^(2) theta -1)) (ii) (sin theta + cos theta ) /(sin theta - cos theta) + ( sin theta - cos theta) /(sin theta + cos theta) = (2) /((1- 2 cos^(2) theta))

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

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

Prove each of the following identities : (i) (1+ cos theta + sin theta)/(1+ cos theta - sin theta) =(1+ sin theta)/(cos theta) (ii) (sin theta + 1- cos theta)/(cos theta - 1 + sin theta) = (1+ sin theta)/(cos theta)

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

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

(sin theta-cos theta)^(2)*(sin theta+cos theta)^(2)=