Is tan^2(x)-sin^2(x)=sin^4(x)sec^2(x) ?

The answer to whether tan^2(x)-sin^2(x)=sin^4(x)sec^2(x) is True

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prove tan^2(x)-sin^2(x)=sin^4(x)sec^2(x)

Solution

prove

tan^{2} (x) - sin^{2} (x) = sin^{4} (x) sec^{2} (x)

Solution

True

Solution steps

tan^{2} (x) - sin^{2} (x) = sin^{4} (x) sec^{2} (x)

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Factor

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

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

Factor

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

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

Factor out common term

sin^{2} (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

sin^{2} (x) sin^{2} (x) = sin^{4} (x)

sin^{2} (x) sin^{2} (x)

Apply exponent rule:

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

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

= sin^{2 + 2} (x)

Add the numbers:

2 + 2 = 4

= sin^{4} (x)

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

cos (x) = \frac{1}{sec ( x )}

\frac{sin ^{4} ( x )}{( \frac{1}{s e c ( x )} ) ^{2}}

Simplify

\frac{sin ^{4} ( x )}{( \frac{1}{s e c ( x )} ) ^{2}}

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

(\frac{1}{sec ( x )})^{2}

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= \frac{sin ^{4} ( x )}{\frac{1}{s e c ^{2} ( x )}}

Apply the fraction rule:

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

= \frac{sin ^{4} ( x ) sec ^{2} ( x )}{1}

Apply rule

\frac{a}{1} = a

= sin^{4} (x) sec^{2} (x)

sin^{4} (x) sec^{2} (x)

sin^{4} (x) sec^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is tan^2(x)-sin^2(x)=sin^4(x)sec^2(x) ?
The answer to whether tan^2(x)-sin^2(x)=sin^4(x)sec^2(x) is True