Is (sec(θ)-cos(θ))/(tan(θ))=sin(θ) ?

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

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prove (sec(θ)-cos(θ))/(tan(θ))=sin(θ)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Join

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

- cos (θ) + \frac{1}{cos ( θ )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

- cos (θ) cos (θ) + 1

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

= - cos^{2} (θ) + 1

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

= - - cos^{2} (θ) + 1

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

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

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

Rewrite using trig identities

\frac{1 - cos ^{2} ( θ )}{sin ( θ )}

Use the Pythagorean identity:

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

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

= \frac{sin ^{2} ( θ )}{sin ( θ )}

Cancel the common factor:

sin (θ)

= sin (θ)

= sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

2 sin (\frac{3 π}{2}) = - 2 + cos (\frac{3 π}{2})

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

sin (x) = \frac{tan ( x ) cot ( x )}{csc ( x )}

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

\frac{cot ^{3} ( t )}{csc ( t )} = cos (t) cot^{2} (t)

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