Is (cos^2(x))/(1-sin(x))=(csc(x)+1)/(csc(x)) ?

The answer to whether (cos^2(x))/(1-sin(x))=(csc(x)+1)/(csc(x)) is True

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Popular Trigonometry >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Simplify

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

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

Cancel

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

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

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

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

Refine

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

Cancel the common factor:

sin (x) - 1

= sin (x) + 1

= sin (x) + 1

= sin (x) + 1

= sin (x) + 1

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

= sin (x) + 1

= sin (x) + 1

= 1 + sin (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is (cos^2(x))/(1-sin(x))=(csc(x)+1)/(csc(x)) ?
The answer to whether (cos^2(x))/(1-sin(x))=(csc(x)+1)/(csc(x)) is True