Solved on Mar 05, 2024

Simplify the given trigonometric expression using fundamental identities and algebraic operations. The simplified expression is $\frac{(cot^2 x - 6)(cot x + 1)}{cot x + 1}$ .

STEP 1

Assumptions
1. We are given the expression $\frac{\cot ^{2} x-5 \cot x-6}{\cot x+1}$ .
2. We will use the fundamental trigonometric identities to simplify the expression.
3. We will perform algebraic operations such as factoring and simplifying fractions where appropriate.

STEP 2

First, we notice that the numerator of the given expression is a quadratic in terms of $\cot x$ . We can attempt to factor this quadratic.
$\cot ^{2} x-5 \cot x-6$

STEP 3

To factor the quadratic, we look for two numbers that multiply to $-6$ (the constant term) and add to $-5$ (the coefficient of the middle term).

STEP 4

The two numbers that satisfy these conditions are $-6$ and $1$ because $(-6) \cdot 1 = -6$ and $(-6) + 1 = -5$ .

STEP 5

We can now write the quadratic as a product of two binomials using these numbers.
$\cot ^{2} x-5 \cot x-6 = (\cot x - 6)(\cot x + 1)$

STEP 6

Now we rewrite the original expression with the factored numerator.
$\frac{\cot ^{2} x-5 \cot x-6}{\cot x+1} = \frac{(\cot x - 6)(\cot x + 1)}{\cot x+1}$

STEP 7

We observe that $\cot x + 1$ appears in both the numerator and the denominator. We can simplify the expression by canceling out the common factor.

STEP 8

After canceling the common factor, we are left with:
$\frac{(\cot x - 6)(\cot x + 1)}{\cot x+1} = \cot x - 6$

STEP 9

The expression is now simplified to:
$\cot x - 6$
This is the simplest form of the given expression using the fundamental identities and appropriate algebraic operations.
The simplified expression is $\cot x - 6$ .

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Simplify the given trigonometric expression using fundamental identities and algebraic operations. The simplified expression is $\frac{(cot^2 x - 6)(cot x + 1)}{cot x + 1}$ .

STEP 1

Assumptions
1. We are given the expression $\frac{\cot ^{2} x-5 \cot x-6}{\cot x+1}$ .
2. We will use the fundamental trigonometric identities to simplify the expression.
3. We will perform algebraic operations such as factoring and simplifying fractions where appropriate.

STEP 2

First, we notice that the numerator of the given expression is a quadratic in terms of $\cot x$ . We can attempt to factor this quadratic.
$\cot ^{2} x-5 \cot x-6$

STEP 3

To factor the quadratic, we look for two numbers that multiply to $-6$ (the constant term) and add to $-5$ (the coefficient of the middle term).

STEP 4

The two numbers that satisfy these conditions are $-6$ and $1$ because $(-6) \cdot 1 = -6$ and $(-6) + 1 = -5$ .

STEP 5

We can now write the quadratic as a product of two binomials using these numbers.
$\cot ^{2} x-5 \cot x-6 = (\cot x - 6)(\cot x + 1)$

STEP 6

Now we rewrite the original expression with the factored numerator.
$\frac{\cot ^{2} x-5 \cot x-6}{\cot x+1} = \frac{(\cot x - 6)(\cot x + 1)}{\cot x+1}$

STEP 7

We observe that $\cot x + 1$ appears in both the numerator and the denominator. We can simplify the expression by canceling out the common factor.

STEP 8

After canceling the common factor, we are left with:
$\frac{(\cot x - 6)(\cot x + 1)}{\cot x+1} = \cot x - 6$

STEP 9

The expression is now simplified to:
$\cot x - 6$
This is the simplest form of the given expression using the fundamental identities and appropriate algebraic operations.
The simplified expression is $\cot x - 6$ .

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Simplify the given trigonometric expression using fundamental identities and algebraic operations. The simplified expression is (cot2x−6)(cotx+1)cotx+1\frac{(cot^2 x - 6)(cot x + 1)}{cot x + 1}cotx+1(cot2x−6)(cotx+1)​.

STEP 1

STEP 2

First, we notice that the numerator of the given expression is a quadratic in terms of cot⁡x\cot xcotx. We can attempt to factor this quadratic.cot⁡2x−5cot⁡x−6\cot ^{2} x-5 \cot x-6cot2x−5cotx−6

STEP 3

To factor the quadratic, we look for two numbers that multiply to −6-6−6 (the constant term) and add to −5-5−5 (the coefficient of the middle term).

STEP 4

The two numbers that satisfy these conditions are −6-6−6 and 111 because (−6)⋅1=−6(-6) \cdot 1 = -6(−6)⋅1=−6 and (−6)+1=−5(-6) + 1 = -5(−6)+1=−5.

STEP 5

We can now write the quadratic as a product of two binomials using these numbers.cot⁡2x−5cot⁡x−6=(cot⁡x−6)(cot⁡x+1)\cot ^{2} x-5 \cot x-6 = (\cot x - 6)(\cot x + 1)cot2x−5cotx−6=(cotx−6)(cotx+1)

STEP 6

Now we rewrite the original expression with the factored numerator.cot⁡2x−5cot⁡x−6cot⁡x+1=(cot⁡x−6)(cot⁡x+1)cot⁡x+1\frac{\cot ^{2} x-5 \cot x-6}{\cot x+1} = \frac{(\cot x - 6)(\cot x + 1)}{\cot x+1}cotx+1cot2x−5cotx−6​=cotx+1(cotx−6)(cotx+1)​

STEP 7

We observe that cot⁡x+1\cot x + 1cotx+1 appears in both the numerator and the denominator. We can simplify the expression by canceling out the common factor.

STEP 8

After canceling the common factor, we are left with:(cot⁡x−6)(cot⁡x+1)cot⁡x+1=cot⁡x−6\frac{(\cot x - 6)(\cot x + 1)}{\cot x+1} = \cot x - 6cotx+1(cotx−6)(cotx+1)​=cotx−6

STEP 9

The expression is now simplified to:cot⁡x−6\cot x - 6cotx−6This is the simplest form of the given expression using the fundamental identities and appropriate algebraic operations.The simplified expression is cot⁡x−6\cot x - 6cotx−6.

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Simplify the given trigonometric expression using fundamental identities and algebraic operations. The simplified expression is $\frac{(cot^2 x - 6)(cot x + 1)}{cot x + 1}$ .

First, we notice that the numerator of the given expression is a quadratic in terms of $\cot x$ . We can attempt to factor this quadratic.
$\cot ^{2} x-5 \cot x-6$

To factor the quadratic, we look for two numbers that multiply to $-6$ (the constant term) and add to $-5$ (the coefficient of the middle term).

The two numbers that satisfy these conditions are $-6$ and $1$ because $(-6) \cdot 1 = -6$ and $(-6) + 1 = -5$ .

We can now write the quadratic as a product of two binomials using these numbers.
$\cot ^{2} x-5 \cot x-6 = (\cot x - 6)(\cot x + 1)$

Now we rewrite the original expression with the factored numerator.
$\frac{\cot ^{2} x-5 \cot x-6}{\cot x+1} = \frac{(\cot x - 6)(\cot x + 1)}{\cot x+1}$

We observe that $\cot x + 1$ appears in both the numerator and the denominator. We can simplify the expression by canceling out the common factor.

After canceling the common factor, we are left with:
$\frac{(\cot x - 6)(\cot x + 1)}{\cot x+1} = \cot x - 6$

The expression is now simplified to:
$\cot x - 6$
This is the simplest form of the given expression using the fundamental identities and appropriate algebraic operations.
The simplified expression is $\cot x - 6$ .