Is (1-sin(θ))/(sec(θ))=(cos^3(θ))/(1+sin(θ)) ?

The answer to whether (1-sin(θ))/(sec(θ))=(cos^3(θ))/(1+sin(θ)) is True

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prove (1-sin(θ))/(sec(θ))=(cos^3(θ))/(1+sin(θ))

prove

\frac{1 - sin ( θ )}{sec ( θ )} = \frac{cos ^{3} ( θ )}{1 + sin ( θ )}

True

Solution steps

\frac{1 - sin ( θ )}{sec ( θ )} = \frac{cos ^{3} ( θ )}{1 + sin ( θ )}

Manipulating left side

\frac{1 - sin ( θ )}{sec ( θ )}

Rewrite using trig identities

Multiply by

\frac{1 + sin ( θ )}{1 + sin ( θ )}

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

Use the Pythagorean identity:

1 = cos^{2} (θ) + sin^{2} (θ)

1 - sin^{2} (θ) = cos^{2} (θ)

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

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

\frac{cos ^{2} ( θ )}{( 1 + sin ( θ ) ) \frac{1}{c o s ( θ )}} : \frac{cos ^{3} ( θ )}{1 + sin ( θ )}

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

Multiply

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

(1 + sin (θ)) \frac{1}{cos ( θ )}

Multiply fractions:

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

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

1 \cdot (1 + sin (θ)) = 1 + sin (θ)

1 \cdot (1 + sin (θ))

Multiply:

1 \cdot (1 + sin (θ)) = (1 + sin (θ))

= (1 + sin (θ))

Remove parentheses:

(a) = a

= 1 + sin (θ)

= \frac{1 + sin ( θ )}{cos ( θ )}

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

Apply the fraction rule:

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

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

cos^{2} (θ) cos (θ) = cos^{3} (θ)

cos^{2} (θ) cos (θ)

Apply exponent rule:

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

cos^{2} (θ) cos (θ) = cos^{2 + 1} (θ)

= cos^{2 + 1} (θ)

Add the numbers:

2 + 1 = 3

= cos^{3} (θ)

= \frac{cos ^{3} ( θ )}{1 + sin ( θ )}

= \frac{cos ^{3} ( θ )}{1 + sin ( θ )}

= \frac{cos ^{3} ( θ )}{1 + sin ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

2 sin (θ) = 2 sin (θ) cos (θ)

5 sin^{2} (x) + 5 cos^{2} (x) = 5

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

\frac{cos ( b )}{sin ( b )} sin (b) = cos (b)

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

Is (1-sin(θ))/(sec(θ))=(cos^3(θ))/(1+sin(θ)) ?
The answer to whether (1-sin(θ))/(sec(θ))=(cos^3(θ))/(1+sin(θ)) is True