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Verify the trigonometric identities

1) (sin x + cos x)^2 + (sin x - cos x)^2=2

2) (2 - cos^2 x) / (sin x) = csc x + sin x

3) (cos x-sin^2 x) / (sin x + cos^2 x) = (csc x - tan x) / (sec x + cot x)

These are the rest of the instructions: For each of the following problems, verify that the left-hand side of the equation is equal to the right-hand side of the equation. Try to simplify the side that is most complex and get it to equal the other side. If you come to a point where you are unable to continue, you may have to start over and simplify the other side instead.
Please show work so I can figure out how to do this problem in the future!

Respuesta :

1)

here, we do the left-hand-side

[tex]\bf [sin(x)+cos(x)]^2+[sin(x)-cos(x)]^2=2 \\\\\\\ [sin^2(x)+2sin(x)cos(x)+cos^2(x)]\\\\+~ [sin^2(x)-2sin(x)cos(x)+cos^2(x)] \\\\\\ 2sin^2(x)+2cos^2(x)\implies 2[sin^2(x)+cos^2(x)]\implies 2[1]\implies 2[/tex]

2)

here we also do the left-hand-side

[tex]\bf \cfrac{2-cos^2(x)}{sin(x)}=csc(x)+sin(x) \\\\\\ \cfrac{2-[1-sin^2(x)]}{sin(x)}\implies \cfrac{2-1+sin^2(x)}{sin(x)}\implies \cfrac{1+sin^2(x)}{sin(x)} \\\\\\ \cfrac{1}{sin(x)}+\cfrac{sin^2(x)}{sin(x)}\implies csc(x)+sin(x)[/tex]

3)

here, we do the right-hand-side

[tex]\bf \cfrac{cos(x)-sin^2(x)}{sin(x)+cos^2(x)}=\cfrac{csc(x)-tan(x)}{sec(x)+cot(x)} \\\\\\ \cfrac{csc(x)-tan(x)}{sec(x)+cot(x)}\implies \cfrac{\frac{1}{sin(x)}-\frac{sin(x)}{cos(x)}}{\frac{1}{cos(x)}-\frac{cos(x)}{sin(x)}}\implies \cfrac{\frac{cos(x)-sin^2(x)}{sin(x)cos(x)}}{\frac{sin(x)+cos^2(x)}{sin(x)cos(x)}} \\\\\\ \cfrac{cos(x)-sin^2(x)}{\underline{sin(x)cos(x)}}\cdot \cfrac{\underline{sin(x)cos(x)}}{sin(x)+cos^2(x)}\implies \cfrac{cos(x)-sin^2(x)}{sin(x)+cos^2(x)}[/tex]

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