Question

$se{c}^6\theta +co{s}^6\theta =1-3si{n}^2\theta co{s}^2\theta$

Answer 1

$LHS=se{c}^6\theta +co{s}^6\theta$

$=(se{c}^2\theta {)}^3+(co{s}^2\theta {)}^3$

$=(se{c}^2\theta +co{s}^2\theta )\left[\right(si{n}^2\theta {)}^2-si{n}^2\theta co{s}^2\theta +\left(co{s}^2\theta {)}^2\right]$

$(\because a^3+b^3=(a+b\left)\right(a^2-ab+b^2\left)\right]$

$=\left[\right(si{n}^2\theta {)}^2+(co{s}^2\theta {)}^2+2si{n}^2\theta co{s}^2\theta -2si{n}^2\theta co{s}^2\theta -si{n}^2\theta co{s}^2\theta ]$

[adding and subtracting $2si{n}^2\theta co{s}^2\theta$ and using identity $si{n}^2\theta +co{s}^2\theta =1]$

$=(si{n}^2\theta +co{s}^2\theta {)}^2-3si{n}^2\theta co{s}^2\theta$

$=(si{n}^2\theta +co{s}^2\theta {)}^2-3si{n}^2\theta co{s}^2\theta$

$=1^2-3si{n}^2\theta co{s}^2\theta$

$(\because si{n}^2\theta +co{s}^2\theta =1)$

$=1-3si{n}^2\theta co{s}^2\theta$

=RHS

$\therefore LHS=RHS$

Hence proved.