sin(theta)^4 %2B 2*(sin(theta)^2*cos(theta)^2) %2B cos(theta)^4=1

1.	sin(theta)^4 %2B 2(sin(theta)^2cos(theta)^2) %2B cos(theta)^4=1
Answer» Let theta = x LHS : sin^4 x + 2sin^2 x cos^2 x + cos^4 x = (sin^2 x)^2 + 2sin^2 x cos^2 x + (cos^2 x)^2 = (sin^2 x + cos^2 x)^2 [Using sin^2 x + cos^2 x = 1] = 1 = RHS Hence proved To prove:sin⁴ theta+2sin²theta cos ²theta+cos ⁴theta =1 LHS:sin⁴ theta+2sin²theta cos ²theta+cos ⁴theta =(sin²theta+cos ²theta)² =(1)² =1

sin(theta)^4 %2B 2(sin(theta)^2cos(theta)^2) %2B cos(theta)^4=1

Answer»

Let theta = x

LHS : sin^4 x + 2sin^2 x cos^2 x + cos^4 x

= (sin^2 x)^2 + 2sin^2 x cos^2 x + (cos^2 x)^2

= (sin^2 x + cos^2 x)^2 [Using sin^2 x + cos^2 x = 1]

= 1

= RHS

Hence proved

To prove:sin⁴ theta+2sin²theta cos ²theta+cos ⁴theta =1

LHS:sin⁴ theta+2sin²theta cos ²theta+cos ⁴theta

=(sin²theta+cos ²theta)²

=(1)²