Question

${\sin }^4\theta -{\cos }^4\theta =1-2{\cos }^2\theta$.

Answer 1

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

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

$=si{n}^2\theta -co{s}^2\theta$ $[\because si{n}^2\theta +co{s}^2\theta =1]$

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

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