Is 4/(cos^2(x))-5=4tan^2(x)-1 ?

The answer to whether 4/(cos^2(x))-5=4tan^2(x)-1 is True

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prove 4/(cos^2(x))-5=4tan^2(x)-1

Solution

prove

\frac{4}{cos ^{2} ( x )} - 5 = 4 tan^{2} (x) - 1

Solution

True

Solution steps

\frac{4}{cos ^{2} ( x )} - 5 = 4 tan^{2} (x) - 1

Manipulating right side

4 tan^{2} (x) - 1

Express with sin, cos

- 1 + 4 tan^{2} (x)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - 1 + 4 (\frac{sin ( x )}{cos ( x )})^{2}

Simplify

- 1 + 4 (\frac{sin ( x )}{cos ( x )})^{2} : \frac{- cos ^{2} ( x ) + 4 sin ^{2} ( x )}{cos ^{2} ( x )}

- 1 + 4 (\frac{sin ( x )}{cos ( x )})^{2}

4 (\frac{sin ( x )}{cos ( x )})^{2} = \frac{4 sin ^{2} ( x )}{cos ^{2} ( x )}

4 (\frac{sin ( x )}{cos ( x )})^{2}

(\frac{sin ( x )}{cos ( x )})^{2} = \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

(\frac{sin ( x )}{cos ( x )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

= 4 \cdot \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{sin ^{2} ( x ) \cdot 4}{cos ^{2} ( x )}

= - 1 + \frac{4 sin ^{2} ( x )}{cos ^{2} ( x )}

Convert element to fraction:

1 = \frac{1 cos ^{2} ( x )}{cos ^{2} ( x )}

= - \frac{1 \cdot cos ^{2} ( x )}{cos ^{2} ( x )} + \frac{sin ^{2} ( x ) \cdot 4}{cos ^{2} ( x )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 1 \cdot cos ^{2} ( x ) + sin ^{2} ( x ) \cdot 4}{cos ^{2} ( x )}

Multiply:

1 \cdot cos^{2} (x) = cos^{2} (x)

= \frac{- cos ^{2} ( x ) + 4 sin ^{2} ( x )}{cos ^{2} ( x )}

= \frac{- cos ^{2} ( x ) + 4 sin ^{2} ( x )}{cos ^{2} ( x )}

= \frac{- cos ^{2} ( x ) + 4 sin ^{2} ( x )}{cos ^{2} ( x )}

Rewrite using trig identities

\frac{- cos ^{2} ( x ) + 4 sin ^{2} ( x )}{cos ^{2} ( x )}

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= \frac{- cos ^{2} ( x ) + 4 ( 1 - cos ^{2} ( x ) )}{cos ^{2} ( x )}

Expand

- cos^{2} (x) + 4 (1 - cos^{2} (x)) : - 5 cos^{2} (x) + 4

- cos^{2} (x) + 4 (1 - cos^{2} (x))

Expand

4 (1 - cos^{2} (x)) : 4 - 4 cos^{2} (x)

4 (1 - cos^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (x)

= 4 \cdot 1 - 4 cos^{2} (x)

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 cos^{2} (x)

= - cos^{2} (x) + 4 - 4 cos^{2} (x)

Simplify

- cos^{2} (x) + 4 - 4 cos^{2} (x) : - 5 cos^{2} (x) + 4

- cos^{2} (x) + 4 - 4 cos^{2} (x)

Group like terms

= - cos^{2} (x) - 4 cos^{2} (x) + 4

Add similar elements:

- cos^{2} (x) - 4 cos^{2} (x) = - 5 cos^{2} (x)

= - 5 cos^{2} (x) + 4

= - 5 cos^{2} (x) + 4

= \frac{- 5 cos ^{2} ( x ) + 4}{cos ^{2} ( x )}

= \frac{- 5 cos ^{2} ( x ) + 4}{cos ^{2} ( x )}

Expand

\frac{4 - 5 cos ^{2} ( x )}{cos ^{2} ( x )} : \frac{4}{cos ^{2} ( x )} - 5

\frac{4 - 5 cos ^{2} ( x )}{cos ^{2} ( x )}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{4 - 5 cos ^{2} ( x )}{cos ^{2} ( x )} = \frac{4}{cos ^{2} ( x )} - \frac{5 cos ^{2} ( x )}{cos ^{2} ( x )}

= \frac{4}{cos ^{2} ( x )} - \frac{5 cos ^{2} ( x )}{cos ^{2} ( x )}

Cancel

\frac{5 cos ^{2} ( x )}{cos ^{2} ( x )} : 5

\frac{5 cos ^{2} ( x )}{cos ^{2} ( x )}

Cancel the common factor:

cos^{2} (x)

= 5

= \frac{4}{cos ^{2} ( x )} - 5

= - 5 + \frac{4}{cos ^{2} ( x )}

= \frac{4}{cos ^{2} ( x )} - 5

We showed that the two sides could take the same form

\Rightarrow True

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Is 4/(cos^2(x))-5=4tan^2(x)-1 ?
The answer to whether 4/(cos^2(x))-5=4tan^2(x)-1 is True