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if [tex]tan (x) = \frac{5}{12}[/tex]
and x is in third quarter
find :
[tex]sin(180+x) + tan(360-x) + \frac{1}{cos(180-x)}[/tex]
Please i want step by step not only the answer

Sagot :

Trancescientidn

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

[tex]\sin x = -\frac{5}{13}\;\;\text{and}\;\;\cos x = -\frac{12}{13}[/tex]

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

Recall the addition identities listed below:

[tex]\sin(\alpha + \beta) = \sin\alpha\sin\beta + \cos\alpha\cos\beta[/tex]

[tex]\Rightarrow \sin(180+x) = \sin180\sin x + \cos180\cos x[/tex]

[tex]\;\;\;\;\;\;= -\sin x = \dfrac{5}{13}[/tex]

[tex]\cos(\alpha - \beta) = \cos\alpha \cos\beta + \sin\alpha \sin\beta[/tex]

[tex]\Rightarrow \cos(180 - x) = \cos180\cos x + \sin180\sin x[/tex]

[tex]\;\;\;\;\;\;=-\cos x = \dfrac{12}{13}[/tex]

[tex]\tan(\alpha - \beta) = \dfrac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}[/tex]

[tex]\Rightarrow \tan(360 - x) = \dfrac{\tan 360 - \tan x}{1 + \tan 360 \tan x}[/tex]

[tex]\;\;\;\;\;\;= -\tan x = -\dfrac{5}{12}[/tex]

Therefore, the expression reduces to

[tex]\sin(180+x) + \tan(360-x) + \frac{1}{\cos(180-x)}[/tex]

[tex]\;\;\;\;\;= \left(\dfrac{5}{13}\right) + \left(\dfrac{5}{12}\right) + \dfrac{1}{\left(\frac{12}{13}\right)}[/tex]

[tex]\;\;\;\;\;= \dfrac{49}{26}[/tex]

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