Question

If $sin\theta +cos\theta =x$, prove that $si{n}^4\theta ÷{\cos }^4\theta =\frac{2-(x^2-1{)}^2}{2}$.

Answer 1

$(sin\theta +cos\theta {)}^2=x^2$

$\Rightarrow 2sin\theta .cos\theta =x^2-1$

$si{n}^4\theta +co{s}^4\theta =(si{n}^2\theta +co{s}^2\theta {)}^2-2si{n}^2\theta .co{s}^2\theta$

$\Rightarrow 1-\frac{(x^2-1{)}^2}{2}$

Hence proved.