Question

hlo que is to prove .........see above attachment...


plz don't post irrelevant answer ✨✨​

Answer 1

$\large \underline \bold{Solution}$:-

$\large \underline \bold{Method \: \: I}$:-

LHS :-

$\: \: \: \: \: \: \: \: \:$$\sf{Sin^{4} \theta \: - \: Cos^{4} \theta}$

$\sf{=> Sin^{4} \theta \: + \: Cos^{4} \theta \: - \: 2Cos^{4} \theta}$

$\sf{=> (Sin^{2} \theta)^{2} \: + \: (Cos^{2} \theta)^{2} \: - \: 2Cos^{4} \theta}$

We know that -

$\: \: \: \: \: \:$$\small\boxed{\sf\red{(a^{2} \: + \: b^{2}) \: = \: (a + b)^{2} \: - \: 2ab}}$

So ,

$\sf{=> (Sin^{2} \theta + Cos^{2} \theta)^{2} \: - \: 2Sin^{2} \theta Cos^{2} \theta \: - \: 2Cos^{4} \theta}$

$\: \: \: \: \: \:$$\small\boxed{\sf\pink{(Sin^{2} \theta \: + \: Cos^{2} \theta) \: = \: 1}}$

then ,

$\sf{=> (1)^{2} \: - \: 2Sin^{2} \theta Cos^{2} \theta \: - \: 2Cos^{4} \theta}$

$\sf{=> 1 \: - \: 2Sin^{2} \theta Cos^{2} \theta \: - \: 2Cos^{4} \theta}$

$\sf{=> 1 \: - \: 2Cos^{2} \theta(Sin^{2} \theta \: + \: Cos^{2} \theta)}$

$\sf{=> 1 \: - \: 2Cos^{2} \theta(1)}$

$\sf{=> 1 \: - \: 2Cos^{2} \theta}$

=>$\: \: \: \: \:$$\small \bold{ RHS}$

$\large \underline \bold{Method \: \: II}$:-

$\: \: \: \: \: \: \: \: \:$$\sf{Sin^{4} \theta \: - \: Cos^{4} \theta}$

$\sf{=> (Sin^{2} \theta)^{2} \: - \: (Cos^{2} \theta)^{2}}$

$\: \: \: \: \: \:$$\small\boxed{\sf\red{(a^{2} \: - \: b^{2}) \: = \: (a \: + \: b)(a \: - \: b)}}$

$\sf{=> (Sin^{2} \theta \: + \: Cos^{2} \theta)(Sin^{2} \theta \: - \: Cos^{2} \theta)}$

$\sf{=> \: \: 1(Sin^{2} \theta \: - \: Cos^{2} \theta)}$

$\sf{=> \: \: \: (Sin^{2} \theta \: - \: Cos^{2} \theta)}$

$\sf{=> \: \: \: 1 \: - \: Cos^{2} \theta \: - \: Cos^{2} \theta}$

$\small \bold{=> \: \: \: \: \: \: 1 \: - \: 2Cos^{2} \theta}$

$\large \underline \bold{HENCE \: PROOF}$