hard trigonometry help?

Question

show work plz.

1. Express cot²x-1/csc²x in terms of sin x.


2. Reduce (csc²x-sec²x) to an expression containing only tan x.


3. Verify the following identities.
a. (sin B + cos B) (sin B – cos B) = 2 sin² B -1

b.(1 – cos² y + sin² y)² +4 sin² y cos² y=4 sin² y
c. tan² 0 sec² 0- sec² 0+1 = tan 0
d. sin 0/csc 0 + cos 0/ sec 0 =1
e. sin x tan ² x cot ^3 x=cos x
f. sec² x/sec ² x-1 =csc² x

Answer 1

1.

(cot^2(x) - 1)/(csc^2(x))

Your first step is to change everything into sine and cosine.
Keep in mind that cot(x) = cos(x)/sin(x), and csc(x) = 1/sin(x).

{[cos(x)/sin(x)]^2 - 1} / {[1/sin(x)]^2}

If we square a fraction, we square the numerator and denominator, therefore:

{ [cos^2(x)/sin^2(x)] - 1} / { 1/sin^2(x) }

Now, in order to get rid of the complex fractions, or "fractions within fractions" we have to multiply both numerator and denominator by sin^2(x). This should leave no more fractions within fractions.

{cos^2(x) - sin^2(x)} / 1

Anything over 1 is itself.

cos^2(x) - sin^2(x)

And since we want to express everything in terms of sinx, we use the identity relating cos^2(x) and sin^2(x). That is, 1 - cos^2(x) = sin^2(x)

[1 - sin^2(x)] - sin^2(x)

We can group like terms.

1 - 2sin^2(x)

---end of question 1---

csc^2(x) - sec^2(x)

Two identities to note:

sec^2(x) = tan^2(x) + 1
csc^2(x) = cot^2(x) + 1

Let's use these.

[cot^2(x) + 1] - [tan^2(x) + 1]

Also, note that cotangent is the reciprocal of tangent. Therefore, coty = 1/tan(y). Let's use this fact.

1/[tan^2(x)] + 1 - tan^2(x) - 1

The 1 and -1 will cancel each other out.

1/[tan^2(x)] - tan^2(x)

3.

LHS = (sin B + cos B) (sin B - cos B)

This is a difference of squares.

sin^2(B) - cos^2(B)

Use the fact that cos^2(x) = 1 - sin^2(x)

sin^2(B) - [1 - sin^2(B)]
2sin^2(B) - 1 = RHS

Answer 2

1. Express cot²x-1/csc²x in terms of sin x.

cot²x - 1/csc²x

= ([csc²x - 1] - 1/csc²x

= 1/sin²x - 1 - 1/csc²x

= 1/sin²x - 1 - sin²x

Just in case you meant (cot²x - 1)/csc²x

(cot²x - 1)/csc²x

= ([csc²x - 1] - 1)/csc²x

= (csc²x - 2)/csc²x

= 1 - 2/csc²x

= 1 - 2sin²x


2. Reduce (csc²x - sec²x) to an expression containing only tan x.

csc²x-sec²x

= [cot²x + 1] - [tan²x + 1]

= 1/tan²x + 1 - tan²x - 1

= 1/tan²x - tan²x

3. Verify the following identities:

a. (sin B + cos B) (sin B – cos B) = 2 sin² B -1

LHS = (sin B + cos B) (sin B – cos B)

= sin²B - cos²B

= sin²B - (1 - sin²B)

= sin²B - 1 + sin²B

= 2sin²B - 1

= RHS .......... QED

b. (1 – cos² y + sin² y)² + 4sin² y cos² y = 4 sin² y

LHS = (1 – cos² y + sin² y)² + 4sin² y cos² y

= (sin² y + sin² y)² + 4sin² y (1 - sin² y)

= 4 sin^4 y + 4 sin² y - 4 sin^4 y

= 4 sin² y

= RHS ...... QED

c. tan² 0 sec² 0 - sec² 0 + 1 = tan 0

LHS = tan² 0 sec² 0 - sec² 0 + 1

= 0 * 1 - 1 + 1

= 0

= tan 0

= RHS ........... QED

d. sin 0/csc 0 + cos 0/sec 0 = 1

LHS = sin 0/csc 0 + cos 0/sec 0

= sin 0 * sin 0 + cos 0 * cos 0

= sin² 0 + cos² 0

= 1

= RHS ..... QED

e. sinx tan²x cot³x = cosx

LHS = sinx tan²x cot³x

= sinx cotx * tan²x cot²x

= sinx * cosx/sinx * tan²x * 1/tan²x

= cosx * 1

= cosx

= RHS ..... QED

f. sec² x/sec² x - 1 =csc² x

Assuming you mean sec² x/(sec ² x - 1) = csc² x (because otherwise sec² x/sec² x - 1 = 1 - 1 = 0)

LHS = sec² x/(sec ² x - 1)

= sec²x/tan²x

= 1/cos²x * cos²x/sin²x

= 1/sin²x

= csc²x

= RHS ..... QED

Answer 3

I took Algebra a million in 8th grade...and that i'm taking Geometry this year...i might think of it relatively is somewhat tricky. i'm taking Algebra 2 next year. Geometry comes after Algebra a million right here.