Is (sec(t)-1)/(tan(t))=(tan(t))/(sec(t)+1) ?

The answer to whether (sec(t)-1)/(tan(t))=(tan(t))/(sec(t)+1) is True

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prove (sec(t)-1)/(tan(t))=(tan(t))/(sec(t)+1)

Solution

prove

\frac{sec ( t ) - 1}{tan ( t )} = \frac{tan ( t )}{sec ( t ) + 1}

Solution

True

Solution steps

\frac{sec ( t ) - 1}{tan ( t )} = \frac{tan ( t )}{sec ( t ) + 1}

Manipulating left side

\frac{sec ( t ) - 1}{tan ( t )}

Express with sin, cos

\frac{- 1 + sec ( t )}{tan ( t )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

= \frac{- 1 + \frac{1}{c o s ( t )}}{\frac{s i n ( t )}{c o s ( t )}}

Simplify

\frac{- 1 + \frac{1}{c o s ( t )}}{\frac{s i n ( t )}{c o s ( t )}} : \frac{- cos ( t ) + 1}{sin ( t )}

\frac{- 1 + \frac{1}{c o s ( t )}}{\frac{s i n ( t )}{c o s ( t )}}

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply:

1 \cdot cos (t) = cos (t)

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (t)

= - - cos (t) + 1

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

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

= \frac{1 - cos ( t )}{sin ( t )}

Rewrite using trig identities

Multiply by

\frac{1 + cos ( t )}{1 + cos ( t )}

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

Expand

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

(1 + cos (t)) (1 - cos (t))

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = 1, b = cos (t)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (t)

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

sin (t)

= \frac{sin ( t )}{1 + cos ( t )}

= \frac{sin ( t )}{1 + cos ( t )}

Manipulating right side

\frac{tan ( t )}{sec ( t ) + 1}

Express with sin, cos

\frac{tan ( t )}{1 + sec ( t )}

Use the basic trigonometric identity:

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

= \frac{\frac{s i n ( t )}{c o s ( t )}}{1 + sec ( t )}

Use the basic trigonometric identity:

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

= \frac{\frac{s i n ( t )}{c o s ( t )}}{1 + \frac{1}{c o s ( t )}}

Simplify

\frac{\frac{s i n ( t )}{c o s ( t )}}{1 + \frac{1}{c o s ( t )}} : \frac{sin ( t )}{cos ( t ) + 1}

\frac{\frac{s i n ( t )}{c o s ( t )}}{1 + \frac{1}{c o s ( t )}}

Apply the fraction rule:

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

= \frac{sin ( t )}{cos ( t ) ( 1 + \frac{1}{c o s ( t )} )}

Join

1 + \frac{1}{cos ( t )} : \frac{cos ( t ) + 1}{cos ( t )}

1 + \frac{1}{cos ( t )}

Convert element to fraction:

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

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

Multiply:

1 \cdot cos (t) = cos (t)

= \frac{cos ( t ) + 1}{cos ( t )}

= \frac{sin ( t )}{\frac{c o s ( t ) + 1}{c o s ( t )} cos ( t )}

Multiply

cos (t) \frac{cos ( t ) + 1}{cos ( t )} : cos (t) + 1

cos (t) \frac{cos ( t ) + 1}{cos ( t )}

Multiply fractions:

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

= \frac{( cos ( t ) + 1 ) cos ( t )}{cos ( t )}

Cancel the common factor:

cos (t)

= cos (t) + 1

= \frac{sin ( t )}{cos ( t ) + 1}

= \frac{sin ( t )}{cos ( t ) + 1}

= \frac{sin ( t )}{1 + cos ( t )}

We showed that the two sides could take the same form

\Rightarrow True

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Is (sec(t)-1)/(tan(t))=(tan(t))/(sec(t)+1) ?
The answer to whether (sec(t)-1)/(tan(t))=(tan(t))/(sec(t)+1) is True