Is (cos(x)cot(x))/(1-sin(x))-1=csc(x) ?

The answer to whether (cos(x)cot(x))/(1-sin(x))-1=csc(x) is True

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prove (cos(x)cot(x))/(1-sin(x))-1=csc(x)

Solution

prove

\frac{cos ( x ) cot ( x )}{1 - sin ( x )} - 1 = csc (x)

Solution

True

Solution steps

\frac{cos ( x ) cot ( x )}{1 - sin ( x )} - 1 = csc (x)

Manipulating left side

\frac{cos ( x ) cot ( x )}{1 - sin ( x )} - 1

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply

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

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

Multiply fractions:

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

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

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

Apply the fraction rule:

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

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

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

Convert element to fraction:

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

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

Multiply:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Apply minus-plus rules

- (- a) = a

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

Simplify

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

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

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

1 \cdot sin (x)

Multiply:

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

= 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)

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

= - sin (x) + 1

= - sin (x) + 1

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

Cancel the common factor:

- sin (x) + 1

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= csc (x)

csc (x)

csc (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is (cos(x)cot(x))/(1-sin(x))-1=csc(x) ?
The answer to whether (cos(x)cot(x))/(1-sin(x))-1=csc(x) is True