Is (sec^2(u)-1)/(sec^2(u))=sin^2(u) ?

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

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prove (sec^2(u)-1)/(sec^2(u))=sin^2(u)

prove

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

True

Solution steps

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

Manipulating left side

\frac{sec ^{2} ( u ) - 1}{sec ^{2} ( u )}

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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} ( u ) + 1}{cos ^{2} ( u )}

Multiply:

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

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (u)

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

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

= 1 - cos^{2} (u)

Rewrite using trig identities

1 - cos^{2} (u)

Use the Pythagorean identity:

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

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

= sin^{2} (u)

= sin^{2} (u)

We showed that the two sides could take the same form

\Rightarrow True

sec (θ) = \frac{tan ( θ )}{csc ( θ )} + cos (θ)

sin (\frac{7 π}{6}) = - \frac{1}{2}

cot (x) + tan (x) = \frac{sec ( x )}{sin ( x )}

4 cos^{2} (x) - 4 sin^{2} (x) = 4 - 8 sin^{2} (x)

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

Is (sec^2(u)-1)/(sec^2(u))=sin^2(u) ?
The answer to whether (sec^2(u)-1)/(sec^2(u))=sin^2(u) is True