Question
Prove that $\sin^4\theta - \cos^4\theta = \sin^2\theta - \cos^2\theta$
$= 2sin^2\theta - 1$
$= 1 - 2 \cos^2\theta$
$= 2sin^2\theta - 1$
$= 1 - 2 \cos^2\theta$
✓
Answer
$L.H.S. = \sin^4\theta - \cos^4\theta$
$L.H.S. = (\sin^2\theta )^2 - (\cos^2\theta )^2$
$L.H.S. = (\sin^2\theta - \cos^2\theta )(\sin^2\theta + \cos^2\theta )$
$L.H.S. = (\sin^2\theta - \cos^2\theta ) x 1$
$L.H.S. = \sin^2\theta - \cos^2\theta$
$L.H.S. = R.H.S.L.H.S.= \sin^2\theta - (1 - \sin^2\theta )$
$L.H.S. = \sin^2\theta - 1 + \sin^2\theta$
$L.H.S. = 2sin^2\theta - 1$
$L.H.S. = R.H.S$
$L.H.S. = 2(1 - \cos^2\theta ) - 1$
$L.H.S. = 2 - 2cos^2\theta - 1$
$L.H.S. = 1 - 2cos^2\theta$
$L.H.S. = R.H.S.$
$L.H.S. = (\sin^2\theta )^2 - (\cos^2\theta )^2$
$L.H.S. = (\sin^2\theta - \cos^2\theta )(\sin^2\theta + \cos^2\theta )$
$L.H.S. = (\sin^2\theta - \cos^2\theta ) x 1$
$L.H.S. = \sin^2\theta - \cos^2\theta$
$L.H.S. = R.H.S.L.H.S.= \sin^2\theta - (1 - \sin^2\theta )$
$L.H.S. = \sin^2\theta - 1 + \sin^2\theta$
$L.H.S. = 2sin^2\theta - 1$
$L.H.S. = R.H.S$
$L.H.S. = 2(1 - \cos^2\theta ) - 1$
$L.H.S. = 2 - 2cos^2\theta - 1$
$L.H.S. = 1 - 2cos^2\theta$
$L.H.S. = R.H.S.$
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