Is cot(x)(sec(x)-cos(x))=sin(x) ?

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

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prove cot(x)(sec(x)-cos(x))=sin(x)

prove

cot (x) (sec (x) - cos (x)) = sin (x)

True

Solution steps

cot (x) (sec (x) - cos (x)) = sin (x)

Manipulating left side

cot (x) (sec (x) - cos (x))

Express with sin, cos

(- cos (x) + sec (x)) cot (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

- cos (x) cos (x) + 1

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

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

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

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

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

Rewrite using trig identities

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

Cancel the common factor:

sin (x)

= sin (x)

= sin (x)

We showed that the two sides could take the same form

\Rightarrow True

tan^{2} (θ) = \frac{1 - cos ( 2 θ )}{1 + cos ( 2 θ )}

cos (x + y) cos (x - y) = cos^{2} (x) - sin^{2} (x)

cos (2 θ) csc^{2} (θ) = csc^{2} (θ) - 2

cos (x) - cos (x) = 0

sec (t) - sin (t) tan (t) = cos (t)

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