If `pi/4 < x < pi/2` then simplify : `sqrt(cos^2x-2cosx.sinx+sin^2x)`

Prove that : cos (pi/4- theta)cos (pi/4-phi)-sin(pi/4-theta) sin (pi/4-phi)= sin (theta+phi) .

If tan (pi//4 + theta) + tan (pi//4 - theta) = 4 , then theta =

(pi)/(4)-(3 pi)/(4)-(pi)/(6)+(pi)/(4)=

cos((pi)/(4)+A) cos((pi)/(4) - B) + sin ((pi)/(4) +A) * sin((pi)/4-B)=

sin(pi)/(4)*sin(3 pi)/(4)*sin(5 pi)/(4)*sin(7 pi)/(4)=

cos ((5pi)/(4)) - sin ((5pi)/(4)) + tan((11pi)/(4)) - cot((11pi)/(4))=

f(x)=min(sinx, cosx),x in (-pi,pi) the values of x where f(x) is non differentiable, is the set A . Then A is subset of (A) {(-3pi)/4, (-pi)/4, (3pi)/4} (B) {(-pi)/4, pi/4, (3pi)/3} (C) {(-3pi)/4, (3pi)/4} (D) {(-3pi)/4, (-pi)/4, pi/4, (3pi)/4}

f(x)=min(sin x,cos x),x in(-pi,pi) the values of x where f(x) is non differentiable,is the set A. Then A is subset of (A) {(-3 pi)/(4),(-pi)/(4),(3 pi)/(4)}(B){(-pi)/(4),(pi)/(4),(3 pi)/(3)}(C){(-3 pi)/(4),(3 pi)/(4)}(D){(-3 pi)/(4),(pi)/(4),(pi)/(4),(3 pi)/(4)}

The solution o the equation cos^(2)theta+s int h eta+1=0 lies in the interval (-pi/4,pi/4) b.(pi/4,3 pi/4)c(3 pi/4,5 pi/4)d.(5 pi/4,7 pi/4)

A solution of the equation cos^(2)theta+s int h eta+1=0 lies in the interval (-pi/4,pi/4) b.(pi/4,3 pi/4)c(3 pi/4,5 pi/4)d*(5 pi/4,7 pi/4)