Is (sec^2(x)-1)/(1+tan^2(x))=sin^2(x) ?

The answer to whether (sec^2(x)-1)/(1+tan^2(x))=sin^2(x) is True

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prove (sec^2(x)-1)/(1+tan^2(x))=sin^2(x)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

\frac{sec ^{2} ( x ) - 1}{1 + tan ^{2} ( x )}

Rewrite using trig identities

\frac{sec ^{2} ( x ) - 1}{1 + tan ^{2} ( x )}

Use the Pythagorean identity:

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

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

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{- 1 + ( \frac{1}{c o s ( x )} ) ^{2}}{( \frac{1}{c o s ( x )} ) ^{2}}

Simplify

\frac{- 1 + ( \frac{1}{c o s ( x )} ) ^{2}}{( \frac{1}{c o s ( x )} ) ^{2}} : - cos^{2} (x) + 1

\frac{- 1 + ( \frac{1}{c o s ( x )} ) ^{2}}{( \frac{1}{c o s ( x )} ) ^{2}}

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= \frac{- 1 + ( \frac{1}{c o s ( x )} ) ^{2}}{\frac{1}{c o s ^{2} ( x )}}

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= \frac{- 1 + \frac{1}{c o s ^{2} ( x )}}{\frac{1}{c o s ^{2} ( x )}}

Apply the fraction rule:

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

= \frac{( - 1 + \frac{1}{c o s ^{2} ( x )} ) cos ^{2} ( x )}{1}

Join

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

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

Multiply:

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

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

= \frac{\frac{- c o s ^{2} ( x ) + 1}{c o s ^{2} ( x )} cos ^{2} ( x )}{1}

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

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

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

= 1 - cos^{2} (x)

Rewrite using trig identities

1 - cos^{2} (x)

Use the Pythagorean identity:

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

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

= sin^{2} (x)

= sin^{2} (x)

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)-1)/(1+tan^2(x))=sin^2(x) ?
The answer to whether (sec^2(x)-1)/(1+tan^2(x))=sin^2(x) is True