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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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

Multiply:

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

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (t)

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

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

= 1 - cos^{2} (t)

Rewrite using trig identities

1 - cos^{2} (t)

Use the Pythagorean identity:

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

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

= sin^{2} (t)

= sin^{2} (t)

We showed that the two sides could take the same form

\Rightarrow True

cos (θ) = cos (- θ)

4 cos^{3} (x) - 3 cos (x) = cos (3 x)

\frac{csc ^{2} ( θ )}{cot ( θ )} = sec (θ) csc (θ)

tan (θ) cot (θ) - cos^{2} (θ) = sin^{2} (θ)

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

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