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Hoochie [10]
3 years ago
5

Simplify 1-cot(x)/tan(x)-1

Mathematics
1 answer:
netineya [11]3 years ago
6 0
\bf  \cfrac{1-cot(x)}{tan(x)-1}\qquad 
\begin{cases}
cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}
\\\\ tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}
\end{cases}\qquad thus
\\\\\\
\cfrac{1-\frac{cos(x)}{sin(x)}}{\frac{sin(x)}{cos(x)}-1}\implies 
\cfrac{\frac{sin(x)-cos(x)}{sin(x)}}{\frac{sin(x)-cos(x)}{cos(x)}}\\\\
-----------------------------\\\\
recall\implies \cfrac{\frac{a}{b}}{\frac{c}{{{ d}}}}\implies \cfrac{a}{b}\cdot \cfrac{{{ d}}}{c}\qquad thus\\\\
-----------------------------

\bf \cfrac{\frac{sin(x)-cos(x)}{sin(x)}}{\frac{sin(x)-cos(x)}{cos(x)}}\implies \cfrac{sin(x)-cos(x)}{sin(x)}\cdot \cfrac{cos(x)}{sin(x)-cos(x)}\implies \boxed{?}
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Answer:

The value of x = -1 makes the denominator of the function equal to zero. That is why this value is not included in the domain of f(x)

Step-by-step explanation:

We have the following expression

f(x) = \frac{x^2+4x+3}{x^2-x-2}

Since the division between zero is not defined then the function f(x) can not include the values of x that make the denominator of the function zero.

Now we search that values of x make 0 the denominator factoring the polynomial x^2-x-2

We need two numbers that when adding them get as a result -1 and when multiplying those numbers, obtain -2 as a result.

These numbers are -2 and 1

Then the factors are:

(x-2) (x + 1)

We do the same with the numerator

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We need two numbers that when adding them get as a result 4 and when multiplying those numbers, obtain 3 as a result.

These numbers are 3 and 1

Then the factors are:

(x+3)(x + 1)

Therefore

f(x) = \frac{(x+3)(x+1)}{(x-2)(x+1)}

Note that \frac{(x+1)}{(x+1)}=1 only if x \neq -1

So since x = -1 is not included in the domain the function has a discontinuity in x = -1

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