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

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

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Popular Trigonometry >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

csc^{2} (x) - csc (x) cot (x)

Express with sin, cos

csc^{2} (x) - cot (x) csc (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

Multiply fractions:

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

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

Multiply:

cos (x) \cdot 1 = cos (x)

= \frac{cos ( 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)

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Cancel

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

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

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

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

Refine

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

Cancel the common factor:

1 - cos (x)

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

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

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

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

= \frac{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 1/(1+cos(x))=csc^2(x)-csc(x)cot(x) ?
The answer to whether 1/(1+cos(x))=csc^2(x)-csc(x)cot(x) is True