Question

sin^6+cos^6+3sin^2cos^2=1
pls every1 try to solve tis

Answer 1

I hope this answer help you

Answer 2

Solution :

Now, LHS = sin⁶θ + cos⁶θ

= (sin²θ)³ + (cos²θ)³

= (sin²θ + cos²θ) {(sin²θ)² + (cos²θ)² - sin²θ cos²θ}

= 1 * {(sin²θ + cos²θ)² - 2sin²θ cos²θ - sin²θ cos²θ}

= 1 - 3 sin²θ cos²θ

= RHS (Proved)