Question

sin^6 x + cos^6 x = 1-3sin^2 x . cos^2 x

sin x . cos x .(1+tan x)(1+cot x)=1+2 sin x .cos x

Answer 1

a: \(=\left(sin^2x+cos^2x\right)^3-3\cdot sin^2x\cdot cos^2x\cdot\left(sin^2x+cos^2x\right)\)

\(=1-3\cdot sin^2x\cdot cos^2x\)

b: \(sinxcosx\cdot\left(1+tanx\right)\left(1+cotx\right)\)

\(=\dfrac{1}{2}sin2x\cdot\left(1+cotx+tanx+1\right)\)

\(=\dfrac{1}{2}sin2x\left(2+\dfrac{1}{tanx}+tanx\right)\)

\(=sin2x+\dfrac{tan^2x+1}{tanx}\cdot\dfrac{1}{2}\cdot sin2x\)

\(=1+2\cdot sinx\cdot cosx\)