Is (sec^2(x)-6tan(x)+7)/(sec^2(x)-5)=(tan(x)-4)/(tan(x)+2) ?

The answer to whether (sec^2(x)-6tan(x)+7)/(sec^2(x)-5)=(tan(x)-4)/(tan(x)+2) is True

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prove (sec^2(x)-6tan(x)+7)/(sec^2(x)-5)=(tan(x)-4)/(tan(x)+2)

Solution

prove

\frac{sec ^{2} ( x ) - 6 tan ( x ) + 7}{sec ^{2} ( x ) - 5} = \frac{tan ( x ) - 4}{tan ( x ) + 2}

Solution

True

Solution steps

\frac{sec ^{2} ( x ) - 6 tan ( x ) + 7}{sec ^{2} ( x ) - 5} = \frac{tan ( x ) - 4}{tan ( x ) + 2}

Manipulating left side

\frac{sec ^{2} ( x ) - 6 tan ( x ) + 7}{sec ^{2} ( x ) - 5}

Rewrite using trig identities

\frac{sec ^{2} ( x ) - 6 tan ( x ) + 7}{sec ^{2} ( x ) - 5}

Use the Pythagorean identity:

sec^{2} (x) = tan^{2} (x) + 1

= \frac{tan ^{2} ( x ) + 1 - 6 tan ( x ) + 7}{tan ^{2} ( x ) + 1 - 5}

Simplify

\frac{tan ^{2} ( x ) + 1 - 6 tan ( x ) + 7}{tan ^{2} ( x ) + 1 - 5} : \frac{tan ( x ) - 4}{tan ( x ) + 2}

\frac{tan ^{2} ( x ) + 1 - 6 tan ( x ) + 7}{tan ^{2} ( x ) + 1 - 5}

tan^{2} (x) + 1 - 6 tan (x) + 7 = tan^{2} (x) - 6 tan (x) + 8

tan^{2} (x) + 1 - 6 tan (x) + 7

Group like terms

= tan^{2} (x) - 6 tan (x) + 1 + 7

Add the numbers:

1 + 7 = 8

= tan^{2} (x) - 6 tan (x) + 8

= \frac{tan ^{2} ( x ) - 6 tan ( x ) + 8}{tan ^{2} ( x ) + 1 - 5}

Add/Subtract the numbers:

1 - 5 = - 4

= \frac{tan ^{2} ( x ) - 6 tan ( x ) + 8}{tan ^{2} ( x ) - 4}

Factor

tan^{2} (x) - 6 tan (x) + 8 : (tan (x) - 2) (tan (x) - 4)

tan^{2} (x) - 6 tan (x) + 8

Break the expression into groups

tan^{2} (x) - 6 tan (x) + 8

Definition

Factors of

8 : 1, 2, 4, 8

8

Divisors (Factors)

Find the Prime factors of

8 : 2, 2, 2

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

Multiply the prime factors of

8 : 4

2 \cdot 2 = 4

4

4

Add the prime factors:

2

Add 1 and the number

8

itself

1, 8

The factors of

8

1, 2, 4, 8

Negative factors of

8 : - 1, - 2, - 4, - 8

Multiply the factors by

- 1

to get the negative factors

- 1, - 2, - 4, - 8

For every two factors such that

u * v = 8,

check if

u + v = - 6

Check

u = 1, v = 8 : u * v = 8, u + v = 9 \Rightarrow

FalseCheck

u = 2, v = 4 : u * v = 8, u + v = 6 \Rightarrow

False

u = - 2, v = - 4

Group into

(a x^{2} + ux) + (vx + c)

(tan^{2} (x) - 2 tan (x)) + (- 4 tan (x) + 8)

= (tan^{2} (x) - 2 tan (x)) + (- 4 tan (x) + 8)

Factor out

tan (x)

from

tan^{2} (x) - 2 tan (x) : tan (x) (tan (x) - 2)

tan^{2} (x) - 2 tan (x)

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

tan^{2} (x) = tan (x) tan (x)

= tan (x) tan (x) - 2 tan (x)

Factor out common term

tan (x)

= tan (x) (tan (x) - 2)

Factor out

- 4

from

- 4 tan (x) + 8 : - 4 (tan (x) - 2)

- 4 tan (x) + 8

Rewrite

8

4 \cdot 2

= - 4 tan (x) + 4 \cdot 2

Factor out common term

- 4

= - 4 (tan (x) - 2)

= tan (x) (tan (x) - 2) - 4 (tan (x) - 2)

Factor out common term

tan (x) - 2

= (tan (x) - 2) (tan (x) - 4)

= \frac{( tan ( x ) - 2 ) ( tan ( x ) - 4 )}{tan ^{2} ( x ) - 4}

Factor

tan^{2} (x) - 4 : (tan (x) + 2) (tan (x) - 2)

tan^{2} (x) - 4

Rewrite

4

2^{2}

= tan^{2} (x) - 2^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

tan^{2} (x) - 2^{2} = (tan (x) + 2) (tan (x) - 2)

= (tan (x) + 2) (tan (x) - 2)

= \frac{( tan ( x ) - 2 ) ( tan ( x ) - 4 )}{( tan ( x ) + 2 ) ( tan ( x ) - 2 )}

Cancel the common factor:

tan (x) - 2

= \frac{tan ( x ) - 4}{tan ( x ) + 2}

= \frac{tan ( x ) - 4}{tan ( x ) + 2}

= \frac{tan ( x ) - 4}{tan ( x ) + 2}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is (sec^2(x)-6tan(x)+7)/(sec^2(x)-5)=(tan(x)-4)/(tan(x)+2) ?
The answer to whether (sec^2(x)-6tan(x)+7)/(sec^2(x)-5)=(tan(x)-4)/(tan(x)+2) is True