" (v) "x sin2y=y cos2x" at "((pi)/(4),(pi)/(2))

Find the equations of tangents and normals to the curve at the point on it: x sin 2y = y cos 2x at ((pi)/4, (pi)/2)

If 2 sin x. cos y=1, then (d ^(2)y)/(dx ^(2)) at ((pi)/(4), (pi)/(4)) is …….

If 2 sin x. cos y=1, then (d ^(2)y)/(dx ^(2)) at ((pi)/(4), (pi)/(4)) is …….

If y=cos^(-1)((1)/(2sin x)) then (d^(2)y)/(dx^(2)) at ((pi)/(4),(pi)/(4)) is

If (sin x)(cos y)=(1)/(2), then (d^(2)y)/(dx^(2)) at ((pi)/(4),(pi)/(4)) is (a)-4(b)-2(c)-6(d)0 at ((pi)/(4),(pi)/(4))

The value of cos y cos((pi)/(2)-x)-cos((pi)/(2)-y)cos x+sin y cos((pi)/(2)-x)+cos x sin((pi)/(2)-y) is zero if (A)x=0(B)y=0(C)x=y(D)n pi+y-(pi)/(4)(n in Z)

Prove that (i) cos (n + 2) x cos (n+1) x +sin (n+2) x sin (n+1) x = cos x) (ii) " cos " .((pi)/(4)-x) " cos " .((pi)/(4)-y) " - sin " ((pi)/(4)-x ) " sin " ((pi)/(4) -y) =" sin " (x+y)

cos ((pi)/(4) - x) cos ((pi)/(4) -y) - sin ((pi)/(4) -x) sin ((pi)/(4) -y) = sin ( x +y)

cos ((pi)/(4) - x) cos ((pi)/(4) -y) - sin ((pi)/(4) -x) sin ((pi)/(4) -y) = sin ( x +y)