Question

Prove that :
${\sin }^{4}A-{\cos }^{4}A={\sin }^{2}A-{\cos }^{2}A=2{\sin }^{2}A-1=1-2{\cos }^{2}A$

Answer 1

We have,
LHS = $si{n}^{4}A$ - $co{s}^{4}A$

$\Rightarrow$ LHS = $(si{n}^{2}A{)}^2-(co{s}^{2}A{)}^2$

$\Rightarrow$ LHS = $(si{n}^{2}A+co{s}^{2}A)(si{n}^{2}A-co{s}^{2}A)$

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

$\Rightarrow$ LHS = $si{n}^{2}A$ $-(1-si{n}^{2}A$ ) = $2si{n}^{2}A-1$

$\Rightarrow$ LHS = $2(1-co{s}^{2}A)-1=1-2co{s}^{2}A$ = RHS