Question

$\frac{{\sin }^4\mathrm{\theta }+{\cos }^4\mathrm{\theta }}{1-2{\sin }^2\mathrm{\theta }{\cos }^2\mathrm{\theta }}=1$

Answer 1

We know
${\sin }^2\theta +{\cos }^2\theta =1$
Squaring on both sides, we get
$$\begin{gathered} {\left({\sin }^2\theta +{\cos }^2\theta \right)}^2=1 \\ \Rightarrow {\left({\sin }^2\theta \right)}^2+{\left({\cos }^2\theta \right)}^2+2{\sin }^2\theta {\cos }^2\theta =1\left[{\left(a+b\right)}^2=a^2+b^2+2ab\right] \\ \Rightarrow {\sin }^4\theta +{\cos }^4\theta =1-2{\sin }^2\theta {\cos }^2\theta \end{gathered}$$
$\therefore \frac{{\sin }^4\theta +{\cos }^4\theta }{1-2{\sin }^2\theta {\cos }^2\theta }=\frac{1-2{\sin }^2\theta {\cos }^2\theta }{1-2{\sin }^2\theta {\cos }^2\theta }=1$