Question

Help me please



I couldn't solve it​

Answer 1

`\text{L.H.S}\\\\=(\sin^4 A - \cos^4 A)/( \sin A + \cos A)\\\\\\=((\sin^2 A)^2-(\cos^2 A)^2)/(\sin A + \cos A)\\\\\\=((\sin^2 A + \cos^2 A)(\sin^2 A-\cos^2 A))/(\sin A + \cos A)\\\\\\=(1\cdot(\sin A+\cos A)(\sin A - \cos A))/(\sin A +\cos A)\\\\\\=\sin A - \cos A\\\\\\=\text{R.H.S}~~~ \\\\\text{Proved.}`

Answer 2

heya! ^^

`\\ \frac{ \sin {}^(4) (A) - \cos {}^(4) (A) }{ (\sin \: A + \cos \: A) } = ( \: sin \: A\: - \: cos \: A \: )\\`

`\\LHS = \frac{ \sin {}^(4) (A) - \cos {}^(4) (A) }{ (\sin \: A + \cos \: A) } \\ \\ \frac{( \sin {}^(2) \: A + \cos {}^(2) \: A)( \sin {}^(2) A - \cos {}^(2) A) }{ (\sin \: A + \cos \: A) } \\ \\ we \: know \: that \: - \: sin {}^(2) A + cos {}^(2) A = 1 \\ \\ \therefore \: \frac{(1)(\sin {}^(2) \: A - \cos {}^(2) \: A)}{(\sin \: A + \cos \: A)}`

now , we're well aware of the algebraic identity -

`a {}^(2) - b {}^(2) = (a + b)(a - b)`

using the identity in the equation above ,

`\dashrightarrow \: \frac{(sin \:A - \: cos \: A)\cancel{(sin \:A + \: cos \: A )}}{\cancel{(sin \: A\: + \: cos \: A)}} \\ \\ \dashrightarrow \: (sin \: A \: - \: cos \: A) = RHS`

hence , proved ~

hope helpful :D