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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

\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 ) )}

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

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

Rewrite using trig identities

\frac{cos ^{2} ( 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 )}

Simplify

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

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

Factor

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

1 - sin^{2} (x)

Factor out common term

- 1

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

Factor

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

sin^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

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

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

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

Factor

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

1 - sin (x)

Factor out common term

- 1

= - (sin (x) - 1)

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

Cancel the common factor:

sin (x) - 1

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

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

= \frac{( \frac{1}{c s c ( x )} + 1 ) csc ( x )}{1}

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot csc (x) = csc (x)

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

= \frac{\frac{c s c ( x ) + 1}{c s c ( x )} csc ( x )}{1}

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

csc (x)

= 1 + csc (x)

1 + csc (x)

1 + csc (x)

= csc (x) + 1

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