What is (x^-2 y^10)^2/x^5 y^-3 using only positive exponents

Mathematics

1 answer:

Eddi Din [679]2 years ago

8 0

Answer:

$\frac{(x^{ - 2}{y}^{10})^{2} }{ {x}^{5} {y}^{ - 3} } \\ \frac{ {x}^{ - 4} {y}^{20} }{ {x}^{5} {y}^{ - 3 } } \\ {x}^{ - 9} {y}^{23} \\ \frac{ {y}^{23} }{ {x}^{9} } \\ ans \: c$

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Use the power reduction formulas to rewrite the expression. (Hint: Your answer should not contain any exponents greater than 1.)

Lapatulllka [165]

Some useful relations and identities:

$\tan x=\dfrac{\sin x}{\cos x}$

$\sin^2x=\dfrac{1-\cos2x}2$

$\cos^2x=\dfrac{1+\cos2x}2$

By the first relation, we have

$\tan^2x\sin^3x=\dfrac{\sin^5x}{\cos^2x}=\dfrac{(\sin^2x)^2\sin x}{\cos^2x}$

Applying the two latter identities, we get

$\dfrac{\left(\frac{1-\cos2x}2\right)^2\sin x}{\frac{1+\cos2x}2}=\dfrac{\frac{1-2\cos2x+\cos^22x}4\sin x}{\frac{1+\cos2x}2}=\dfrac{(1-2\cos2x+\cos^22x)\sin x}{2(1+\cos2x)}$

We can apply the third identity again:

$\dfrac{(1-2\cos2x+\cos^22x)\sin x}{2(1+\cos2x)}=\dfrac{\left(1-2\cos2x+\frac{1+\cos4x}2\right)\sin x}{2(1+\cos2x)}=\dfrac{(3-4\cos2x+\cos4x)\sin x}{4(1+\cos2x)}$