First, we need to find the values of the sine and cosine of x knowing the value of tan x and x being in the 3rd quadrant. Since tan x = 5/12, using Pythagorean theorem, we know that

$\sin x = -\frac{5}{13}\;\;\text{and}\;\;\cos x = -\frac{12}{13}$

Note that both sine and cosine are negative because x is in the 3rd quadrant.

Recall the addition identities listed below:

$\sin(\alpha + \beta) = \sin\alpha\sin\beta + \cos\alpha\cos\beta$

$\Rightarrow \sin(180+x) = \sin180\sin x + \cos180\cos x$

$\;\;\;\;\;\;= -\sin x = \dfrac{5}{13}$

$\cos(\alpha - \beta) = \cos\alpha \cos\beta + \sin\alpha \sin\beta$

$\Rightarrow \cos(180 - x) = \cos180\cos x + \sin180\sin x$

$\;\;\;\;\;\;=-\cos x = \dfrac{12}{13}$

$\tan(\alpha - \beta) = \dfrac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}$

$\Rightarrow \tan(360 - x) = \dfrac{\tan 360 - \tan x}{1 + \tan 360 \tan x}$

$\;\;\;\;\;\;= -\tan x = -\dfrac{5}{12}$

Therefore, the expression reduces to

$\sin(180+x) + \tan(360-x) + \frac{1}{\cos(180-x)}$

$\;\;\;\;\;= \left(\dfrac{5}{13}\right) + \left(\dfrac{5}{12}\right) + \dfrac{1}{\left(\frac{12}{13}\right)}$

$\;\;\;\;\;= \dfrac{49}{26}$

5 0

3 years ago

I NEED HELP ASAP thank you so much!!

egoroff_w [7]

B is correct

good luck

5 0

3 years ago

Help please and show work

Lady bird [3.3K]

So the radius is 7 inch which is given
Use the formula for the circumference which is 2xpiXradius
2x3.1415x7=44 to nearest integer

8 0

3 years ago

5x + 2y = 7 y = x + 1 What is the solution set of the given system? {(5/7, 12/7)} {(12/7, 5/7)} {(9/7, 16/7)} {(16/7, 9/7)}

Rama09 [41]

5x+2y=7

y=x+1

5x+2(x+1) =7

5x +2x+2=7

7x +2 =7

7x = 5

x = 5/7

y=5/7 +1

y =12/7

(5/7, 12/7)

5 0

3 years ago