Is (sec(x)-cos(x))/(tan(x))=sin(x) ?

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

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prove (sec(x)-cos(x))/(tan(x))=sin(x)

prove

\frac{sec ( x ) - cos ( x )}{tan ( x )} = sin (x)

True

Solution steps

\frac{sec ( x ) - cos ( x )}{tan ( x )} = sin (x)

Manipulating left side

\frac{sec ( x ) - cos ( x )}{tan ( x )}

Express with sin, cos

\frac{- cos ( x ) + sec ( x )}{tan ( x )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

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

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

sin (2 x) = 2 cos (x) sin (x)

cos (x + π) = - cos (x)

sin (x) sec (x) cot (x) = 1

tan (2 u) = \frac{2 cot ( u )}{csc ^{2} ( u ) - 2}

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

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