Respuesta :

loneguy

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

MidnightShowers

Answer:

heya! ^^

[tex] \\ \frac{ \sin {}^{4} (A) - \cos {}^{4} (A) }{ (\sin \: A + \cos \: A) } = ( \: sin \: A\: - \: cos \: A \: )\\[/tex]

[tex] \\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)} [/tex]

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

[tex]a {}^{2} - b {}^{2} = (a + b)(a - b)[/tex]

using the identity in the equation above ,

[tex]\dashrightarrow \: \frac{(sin \:A - \: cos \: A)\cancel{(sin \:A + \: cos \: A )}}{\cancel{(sin \: A\: + \: cos \: A)}} \\ \\ \dashrightarrow \: (sin \: A \: - \: cos \: A) = RHS[/tex]

hence , proved ~

hope helpful :D

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