Is cos(θ)=(sec(θ)+tan(θ))(1-sin(θ)) ?

The answer to whether cos(θ)=(sec(θ)+tan(θ))(1-sin(θ)) is True

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prove cos(θ)=(sec(θ)+tan(θ))(1-sin(θ))

prove

cos (θ) = (sec (θ) + tan (θ)) (1 - sin (θ))

True

Solution steps

cos (θ) = (sec (θ) + tan (θ)) (1 - sin (θ))

Manipulating right side

(sec (θ) + tan (θ)) (1 - sin (θ))

Express with sin, cos

(1 - sin (θ)) (sec (θ) + tan (θ))

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Simplify

\frac{1}{cos ( θ )} + \frac{sin ( θ )}{cos ( θ )} : \frac{sin ( θ ) + 1}{cos ( θ )}

\frac{1}{cos ( θ )} + \frac{sin ( θ )}{cos ( θ )}

Apply rule

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

= \frac{1 + sin ( θ )}{cos ( θ )}

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Expand

(1 + sin (θ)) (1 - sin (θ)) : 1 - sin^{2} (θ)

(1 + sin (θ)) (1 - sin (θ))

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = 1, b = sin (θ)

= 1^{2} - sin^{2} (θ)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (θ)

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

Use the Pythagorean identity:

1 = cos^{2} (θ) + sin^{2} (θ)

1 - sin^{2} (θ) = cos^{2} (θ)

= \frac{cos ^{2} ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= cos (θ)

= cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

sec (2 x) = \frac{1}{cos ( 2 x )}

\frac{sin ( 6 x ) + sin ( 2 x )}{cos ( 6 x ) - cos ( 2 x )} = - cot (2 x)

(sin (x) + cos (x)) 2 = 1 + sin (2 x)

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

Is cos(θ)=(sec(θ)+tan(θ))(1-sin(θ)) ?
The answer to whether cos(θ)=(sec(θ)+tan(θ))(1-sin(θ)) is True