Is cot(x)-cot(x)cos^2(x)=sin(x)cos(x) ?

The answer to whether cot(x)-cot(x)cos^2(x)=sin(x)cos(x) is True

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prove cot(x)-cot(x)cos^2(x)=sin(x)cos(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

\frac{cos ( x )}{sin ( x )} - cos^{2} (x) \frac{cos ( x )}{sin ( x )} : \frac{cos ( x ) - cos ^{3} ( x )}{sin ( x )}

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

cos^{2} (x) \frac{cos ( x )}{sin ( x )} = \frac{cos ^{3} ( x )}{sin ( x )}

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

Multiply fractions:

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

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

cos (x) cos^{2} (x) = cos^{3} (x)

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

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= cos^{1 + 2} (x)

Add the numbers:

1 + 2 = 3

= cos^{3} (x)

= \frac{cos ^{3} ( x )}{sin ( x )}

= \frac{cos ( x )}{sin ( x )} - \frac{cos ^{3} ( x )}{sin ( x )}

Apply rule

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

= \frac{cos ( x ) - cos ^{3} ( x )}{sin ( x )}

= \frac{cos ( x ) - cos ^{3} ( x )}{sin ( x )}

= \frac{cos ( x ) - cos ^{3} ( x )}{sin ( x )}

Factor

\frac{cos ( x ) - cos ^{3} ( x )}{sin ( x )} : \frac{cos ( x ) ( 1 - cos ^{2} ( x ) )}{sin ( x )}

\frac{cos ( x ) - cos ^{3} ( x )}{sin ( x )}

Factor

cos (x) - cos^{3} (x) : cos (x) (1 - cos^{2} (x))

cos (x) - cos^{3} (x)

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

cos^{3} (x) = cos (x) cos^{2} (x)

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

Factor out common term

cos (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

sin (x)

= sin (x) cos (x)

= sin (x) cos (x)

We showed that the two sides could take the same form

\Rightarrow True

cot^{2} (θ) = \frac{csc ^{2} ( θ )}{sec ^{2} ( θ )}

\frac{1}{sec ( θ )} = arcsec (θ)

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

\frac{1 + cot ^{2} ( x )}{cot ^{2} ( x )} = sec^{2} (x)

sec^{2} (x) - tan^{2} (x) = sec (x) cos (x)

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