Let `f(x)=|1+sin^2x cos^2x4s in2xsin^2\ \ x1+cos^2x4s in2xsin^2xcos^2x1+4s in2x|,\ ` then the maximum value of `f(x)=` a.`2` b. `4` c.`6` d. `8`

Prove that : cos4x=1-8sin^2xcos^2x

Solve: 4sinxsin2xsin4x=sin3x

(i) Find maximum value of f(x)=|{:(1+sin^(2)x,cos^(2)x,4sin2x),(sin^(2)x,1+cos^(2)x,4sin2x),(sin^(2)x,cos^(2)x,1+4sin2x):}| . (ii) Let A,B and C be the angles of triangle such that Agt=Bgt=C. Find the minimum value of Delta where Delta=|{:(sin^(2)A,sinAcosA,cos^(2)A),(sin^(2) B,sinBcosB,cos^(2)B),(sin^(2)C,sinCcosC,cos^(2)C):}| .

Let P(x)=((1-cos2x+sin2x)/(1+cos2x+s in2x))^2+((1+cotx+cot^2x)/(1+tanx+tan^2x)), then the minimum value of P(x) equal 1 (b) 2 (c) 4 (d) 16

Let P(x)=((1-cos2x+sin2x)/(1+cos2x+sin2x))^2+((1+cotx+cot^2x)/(1+tanx+tan^2x)), then the minimum value of P(x) equal 1 (b) 2 (c) 4 (d) 16

The integral int (sin^2xcos^2x)/(sin^5x+cos^3xsin^2x+sin^3xcos^2x+cos^5x)^2 dx is equal to

If sin x+sin^2=1 , then cos^8 x+2 cos^6 x+cos^4 x=

int(sin^8x-cos^8x)/(1-2sin^2xcos^2x)dx=

Evaluate: int(sin^8x-cos^8x)/(1-2sin^2xcos^2x)\ dx

int(sin^8x-cos^8x)/(1-2sin^2xcos^2x)dx=