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

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

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

prove

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

True

Solution steps

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

Manipulating left side

(1 - cos (θ)) (1 + sec (θ))

Express with sin, cos

(1 + sec (θ)) (1 - cos (θ))

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

1 + \frac{1}{cos ( θ )}

Convert element to fraction:

1 = \frac{1 cos ( θ )}{cos ( θ )}

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

Multiply:

1 \cdot cos (θ) = cos (θ)

= \frac{cos ( θ ) + 1}{cos ( θ )}

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Expand

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

(1 + cos (θ)) (1 - cos (θ))

Apply Difference of Two Squares Formula:

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

a = 1, b = cos (θ)

= 1^{2} - cos^{2} (θ)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (θ)

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

Use the Pythagorean identity:

1 = cos^{2} (θ) + sin^{2} (θ)

1 - cos^{2} (θ) = sin^{2} (θ)

= \frac{sin ^{2} ( θ )}{cos ( θ )}

= \frac{sin ^{2} ( θ )}{cos ( θ )}

Rewrite using trig identities

= sin (θ) \frac{sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

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

sin (θ) tan (θ)

sin (θ) tan (θ)

We showed that the two sides could take the same form

\Rightarrow True

tan (y) - \frac{sec ^{2} ( y )}{tan ( y )} = - cot (y)

(csc (x) - cot (x)) (sec (x) 1) = tan (x)

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

tan (2 t) = 2 sin (t) cos (t) sec (2 t)

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

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