Solutions for Precalculus Math Problems?

Question

Hi,

Please provide solutions with step by work for each problem.

1) Simplify: 1/3x + (x ² +1)/(x ² -4x) - (x-2)/(9x-36)

2) Simplify: [(3/4)-(5/8)]/(3/8)

3) Solve: 1/3x + (x ² +1)/(x ² -4x) - (x-2)/(9x-36) = 1/3

4) Verify the identity: tan x(cot x + tan x)=sec ² x

5)Solve graphically: x ² -2x-6=8√(2x+7)

Thanks.

Answer 1

Do your homework.

Answer 2

i only know #2
3/4-5/8=1/8
1/8 / 3/8 =1/3

Answer 3

nicely, if (x-2) squared [we are able to write this as (x-2)^2] = 2, Then x-2 might desire to be sq. root of two, provided that sq. root of (x-2)^2 is x-2. yet, minus the sq. root of two accelerated via minus sq. root of two additionally provides 2, So x-2 = +or- root2 x = 2+root2 or 2=root2

Answer 4

4) tan(x) [cot(x) + tan(x)] = sec^2(x)

Choosing the left hand side (LHS):

LHS = tan(x) [cot(x) + tan(x)]

Change everything to sines and cosines.

LHS = [sin(x)/cos(x)] [cos(x)/sin(x) + sin(x)/cos(x)]

Making a common denominator in the second set of brackets,

LHS = [sin(x)/cos(x)] [ (cos^2(x) + sin^2(x)) / (sin(x)cos(x)) ]

Using the identity that sine squared plus cosines squares equals 1.

LHS = [sin(x)/cos(x)] [1 / (sin(x)cos(x))]

Merging into one fraction,

LHS = [sin(x)/(sin(x)cos^2(x))]

Cancelling out sin(x),

LHS = [1 / cos^2(x)]

And, by definition,

LHS = sec^2(x) = RHS