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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

csc (t) + cot (t)

Express with sin, cos

cot (t) + csc (t)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

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

Simplify

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

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

Multiply by

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

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

Apply Difference of Two Squares Formula:

t^{2} - y^{2} = (t + y) (t - y)

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

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

Use the Pythagorean identity:

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

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

= \frac{sin ( t )}{\frac{s i n ^{2} ( t )}{1 + c o s ( t )}}

Simplify

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

\frac{sin ( t )}{\frac{s i n ^{2} ( t )}{1 + c o s ( t )}}

Apply the fraction rule:

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

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

Cancel the common factor:

sin (t)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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