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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

\frac{sin ( x )}{csc ( x ) - cot ( x )}

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Combine the fractions

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

Apply rule

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

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

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

Apply the fraction rule:

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

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

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 ) + 1}

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

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

Rewrite using trig identities

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

Simplify

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

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

Cancel

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

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

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

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

Refine

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

Cancel the common factor:

cos (x) - 1

= cos (x) + 1

= cos (x) + 1

= cos (x) + 1

= cos (x) + 1

= 1 + cos (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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