Is 1/(csc(x)-1)+1/(csc(x)+1)=2tan(x)sec(x) ?

The answer to whether 1/(csc(x)-1)+1/(csc(x)+1)=2tan(x)sec(x) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{1}{- 1 + \frac{1}{s i n ( x )}} + \frac{1}{1 + \frac{1}{s i n ( x )}}

Simplify

\frac{1}{- 1 + \frac{1}{s i n ( x )}} + \frac{1}{1 + \frac{1}{s i n ( x )}} : \frac{2 sin ( x )}{( sin ( x ) + 1 ) ( - sin ( x ) + 1 )}

\frac{1}{- 1 + \frac{1}{s i n ( x )}} + \frac{1}{1 + \frac{1}{s i n ( x )}}

\frac{1}{- 1 + \frac{1}{s i n ( x )}} = \frac{sin ( x )}{- sin ( x ) + 1}

\frac{1}{- 1 + \frac{1}{s i n ( x )}}

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

= \frac{1}{\frac{- s i n ( x ) + 1}{s i n ( x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

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

\frac{1}{1 + \frac{1}{s i n ( x )}}

Join

1 + \frac{1}{sin ( x )} : \frac{sin ( x ) + 1}{sin ( x )}

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

Convert element to fraction:

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

= \frac{1 \cdot sin ( x )}{sin ( x )} + \frac{1}{sin ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot sin ( x ) + 1}{sin ( x )}

Multiply:

1 \cdot sin (x) = sin (x)

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

= \frac{1}{\frac{s i n ( x ) + 1}{s i n ( x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

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

Least Common Multiplier of

- sin (x) + 1, sin (x) + 1 : (sin (x) + 1) (- sin (x) + 1)

- sin (x) + 1, sin (x) + 1

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

- sin (x) + 1

sin (x) + 1

= (sin (x) + 1) (- sin (x) + 1)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

(sin (x) + 1) (- sin (x) + 1)

For

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

multiply the denominator and numerator by

sin (x) + 1

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

For

\frac{sin ( x )}{sin ( x ) + 1} :

multiply the denominator and numerator by

- sin (x) + 1

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

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

Since the denominators are equal, combine the fractions:

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

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

Expand

sin (x) (sin (x) + 1) + sin (x) (- sin (x) + 1) : 2 sin (x)

sin (x) (sin (x) + 1) + sin (x) (- sin (x) + 1)

Expand

sin (x) (sin (x) + 1) : sin^{2} (x) + sin (x)

sin (x) (sin (x) + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = sin (x), b = sin (x), c = 1

= sin (x) sin (x) + sin (x) \cdot 1

= sin (x) sin (x) + 1 \cdot sin (x)

Simplify

sin (x) sin (x) + 1 \cdot sin (x) : sin^{2} (x) + sin (x)

sin (x) sin (x) + 1 \cdot sin (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)

1 \cdot sin (x) = sin (x)

1 \cdot sin (x)

Multiply:

1 \cdot sin (x) = sin (x)

= sin (x)

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

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

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

Expand

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

sin (x) (- sin (x) + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = sin (x), b = - sin (x), c = 1

= sin (x) (- sin (x)) + sin (x) \cdot 1

Apply minus-plus rules

+ (- a) = - a

= - sin (x) sin (x) + 1 \cdot sin (x)

Simplify

- sin (x) sin (x) + 1 \cdot sin (x) : - sin^{2} (x) + sin (x)

- sin (x) sin (x) + 1 \cdot sin (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)

1 \cdot sin (x) = sin (x)

1 \cdot sin (x)

Multiply:

1 \cdot sin (x) = sin (x)

= sin (x)

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

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

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

Simplify

sin^{2} (x) + sin (x) - sin^{2} (x) + sin (x) : 2 sin (x)

sin^{2} (x) + sin (x) - sin^{2} (x) + sin (x)

Add similar elements:

sin^{2} (x) - sin^{2} (x) = 0

= sin (x) + sin (x)

Add similar elements:

sin (x) + sin (x) = 2 sin (x)

= 2 sin (x)

= 2 sin (x)

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

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

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

Rewrite using trig identities

\frac{2 sin ( x )}{( 1 + sin ( x ) ) ( 1 - sin ( x ) )}

Expand

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

(1 + sin (x)) (1 - sin (x))

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (x)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (x)

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

Use the Pythagorean identity:

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

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

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

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

Manipulating right side

2 tan (x) sec (x)

Express with sin, cos

2 sec (x) tan (x)

Use the basic trigonometric identity:

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

= 2 \cdot \frac{1}{cos ( x )} tan (x)

Use the basic trigonometric identity:

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

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

Simplify

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

2 \cdot \frac{1}{cos ( x )} \cdot \frac{sin ( x )}{cos ( x )}

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

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

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove (1+cos(t))/(1-cos(t))=(csc(t)+cot(t))^2 prove

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

prove sec^2(α)+csc^2(α)= 1/(sin^2(α)cos^2(α))prove

sec^{2} (α) + csc^{2} (α) = \frac{1}{sin ^{2} ( α ) cos ^{2} ( α )}

prove sin^2(A)+cos^2(A)=1 prove

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

prove tan(x)=sin(x)sec(x)prove

tan (x) = sin (x) sec (x)

prove (1+cot^2(θ))(1-cos^2(θ))=1 prove

(1 + cot^{2} (θ)) (1 - cos^{2} (θ)) = 1

Frequently Asked Questions (FAQ)

Is 1/(csc(x)-1)+1/(csc(x)+1)=2tan(x)sec(x) ?
The answer to whether 1/(csc(x)-1)+1/(csc(x)+1)=2tan(x)sec(x) is True