Can SOMEONE REAL ACTUALLY HELP ME LIKE DANG

Verify that:

Lelu [443]

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to verify that:

$\displaystyle \left(\cos(x)\right)\left(\cot(x)\right)=\csc(x)-\sin(x)$

Note that cot(x) = cos(x) / sin(x). Hence:

$\displaystyle \left(\cos(x)\right)\left(\frac{\cos(x)}{\sin(x)}\right)=\csc(x)-\sin(x)$

Multiply:

$\displaystyle \frac{\cos^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Recall that Pythagorean Identity: sin²(x) + cos²(x) = 1 or cos²(x) = 1 - sin²(x). Substitute:

$\displaystyle \frac{1-\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Split:

$\displaystyle \frac{1}{\sin(x)}-\frac{\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Simplify:

$\csc(x)-\sin(x)=\csc(x)-\sin(x)$

Problem 2)

We want to verify that:

$\displaystyle (\csc(x)-\cot(x))^2=\frac{1-\cos(x)}{1+\cos(x)}$

Square:

$\displaystyle \csc^2(x)-2\csc(x)\cot(x)+\cot^2(x)=\frac{1-\cos(x)}{1+\cos(x)}$

Convert csc(x) to 1 / sin(x) and cot(x) to cos(x) / sin(x). Thus:

$\displaystyle \frac{1}{\sin^2(x)}-\frac{2\cos(x)}{\sin^2(x)}+\frac{\cos^2(x)}{\sin^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor out the sin²(x) from the denominator:

$\displaystyle \frac{1}{\sin^2(x)}\left(1-2\cos(x)+\cos^2(x)\right)=\frac{1-\cos(x)}{1+\cos(x)}$

Factor (perfect square trinomial):

$\displaystyle \frac{1}{\sin^2(x)}\left((\cos(x)-1)^2\right)=\frac{1-\cos(x)}{1+\cos(x)}$

Using the Pythagorean Identity, we know that sin²(x) = 1 - cos²(x). Hence:

$\displaystyle \frac{(\cos(x)-1)^2}{1-\cos^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor (difference of two squares):

$\displaystyle \frac{(\cos(x)-1)^2}{(1-\cos(x))(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor out a negative from the first factor in the denominator:

$\displaystyle \frac{(\cos(x)-1)^2}{-(\cos(x)-1)(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Cancel:

$\displaystyle \frac{\cos(x)-1}{-(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Distribute the negative into the numerator. Therefore:

$\displaystyle \frac{1-\cos(x)}{1+\cos(x)}=\displaystyle \frac{1-\cos(x)}{1+\cos(x)}$

3 0

3 years ago

Write an equation perpendicular to the line 6x+12y=24 and passing through (4,0)

noname [10]

Y = 2x - 8 is perpendicular to 6x+12y=24 and passes through (4,0)

6 0

3 years ago

Is the relation a function?

alexira [117]

Answer: yes its just set up diffrently

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Please answer and i have to show work

stealth61 [152]

Answer:

B

Step-by-step explanation:

Calculate the slope m using the slope formula and equate to $\frac{3}{2}$

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (3, 2) and (x₂, y₂ ) = (r, - 4)

m = $\frac{-4-2}{r-3}$ = $\frac{-6}{r-3}$ , then equating gives

$\frac{-6}{r-3}$ = $\frac{3}{2}$ ( cross- multiply )

3(r - 3) = - 12 ( divide both sides by 3 )

r - 3 = - 4 ( add 3 to both sides )

r = - 1 → B

4 0

3 years ago

Ǫᴜɪᴢ-<br><br> What is <img src="https://tex.z-dn.net/?f=x%20%3D%20-12" id="TexFormula1" title="x = -12" alt="x = -12" align="abs

Pachacha [2.7K]

Answer:

x = -12

is x is equal to - twelve

5 0

3 years ago

Read 2 more answers