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

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

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prove cos(x)+cos(x)cot^2(x)=cot(x)csc(x)

Solution

prove

cos (x) + cos (x) cot^{2} (x) = cot (x) csc (x)

Solution

True

Solution steps

cos (x) + cos (x) cot^{2} (x) = cot (x) csc (x)

Manipulating left side

cos (x) + cos (x) cot^{2} (x)

Factor

cos (x) + cos (x) cot^{2} (x) : cos (x) (1 + cot^{2} (x))

cos (x) + cos (x) cot^{2} (x)

Factor out common term

cos (x)

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

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

Express with sin, cos

(1 + cot^{2} (x)) cos (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply:

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

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

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Simplify

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

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

Manipulating right side

cot (x) csc (x)

Express with sin, cos

cot (x) csc (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove tan(a)=tan(a)csc^2(a)+cot(-a)prove

tan (a) = tan (a) csc^{2} (a) + cot (- a)

prove 7csc^2(x)-5cot^2(x)=2csc^2(x)+5 prove

7 csc^{2} (x) - 5 cot^{2} (x) = 2 csc^{2} (x) + 5

prove cos^2(x)(1+cot^2(x))=cot^2(x)prove

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

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

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

prove tan(θ)(cot(θ)+tan(θ))=sec^2(θ)prove

tan (θ) (cot (θ) + tan (θ)) = sec^{2} (θ)

Frequently Asked Questions (FAQ)

Is cos(x)+cos(x)cot^2(x)=cot(x)csc(x) ?
The answer to whether cos(x)+cos(x)cot^2(x)=cot(x)csc(x) is True