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Papessa [141]
2 years ago
10

Help me please I couldn't solve it​

Mathematics
2 answers:
sveta [45]2 years ago
5 0

Answer:

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

inysia [295]2 years ago
4 0

\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.}

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