Question

If }" alt="tan (x) = \frac{5}{12}" align="absmiddle" class="latex-formula">
and x is in third quarter
find :
$sin(180+x) + tan(360-x) + \frac{1}{cos(180-x)}$
Please i want step by step not only the answer

Answer 1

Explanation:

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

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

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

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