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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

\frac{sin ( x ) tan ( x )}{cos ( x ) + 1}

Express with sin, cos

\frac{sin ( x ) tan ( x )}{1 + cos ( x )}

Use the basic trigonometric identity:

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

= \frac{sin ( x ) \frac{s i n ( x )}{c o s ( x )}}{1 + cos ( x )}

Simplify

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

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

Multiply

sin (x) \frac{sin ( x )}{cos ( x )} : \frac{sin ^{2} ( x )}{cos ( x )}

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

Multiply fractions:

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

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

sin (x) sin (x) = sin^{2} (x)

sin (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

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

1 - cos^{2} (x)

Factor out common term

- 1

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

Factor

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

cos^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

= (cos (x) + 1) (cos (x) - 1)

= - (cos (x) + 1) (cos (x) - 1)

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

Cancel the common factor:

cos (x) + 1

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

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

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

Expand

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

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

Apply the fraction rule:

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

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

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

Remove parentheses:

(a) = a, - (- a) = a

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

Apply rule

\frac{a}{a} = 1

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

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

Simplify

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

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

Convert element to fraction:

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

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

Multiply:

1 \cdot cos (x) = cos (x)

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1 - \frac{1}{s e c ( x )}}{\frac{1}{s e c ( x )}}

Simplify

\frac{1 - \frac{1}{s e c ( x )}}{\frac{1}{s e c ( x )}}

Apply the fraction rule:

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

= \frac{( 1 - \frac{1}{s e c ( x )} ) sec ( x )}{1}

Join

1 - \frac{1}{sec ( x )} : \frac{sec ( x ) - 1}{sec ( x )}

1 - \frac{1}{sec ( x )}

Convert element to fraction:

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

= \frac{1 \cdot sec ( x )}{sec ( x )} - \frac{1}{sec ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot sec ( x ) - 1}{sec ( x )}

Multiply:

1 \cdot sec (x) = sec (x)

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

= \frac{\frac{s e c ( x ) - 1}{s e c ( x )} sec ( x )}{1}

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

sec (x)

= sec (x) - 1

sec (x) - 1

sec (x) - 1

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove sin(a+b)-sin(a-b)=2sin(a)sin(b)prove

sin (a + b) - sin (a - b) = 2 sin (a) sin (b)

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

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

prove 1/(cot^2(x))+sec(x)cos(x)=csc^2(x)prove

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

prove (sin(x))/(1-cos^2(x))=cos(x)prove

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

prove 1csc(θ)-(cos(θ))(cot(θ))=sin(θ)prove

1 csc (θ) - (cos (θ)) (cot (θ)) = sin (θ)

Frequently Asked Questions (FAQ)

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