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

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

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

prove

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

True

Solution steps

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

Manipulating left side

\frac{1 - sin ( x )}{sec ( x )}

Rewrite using trig identities

Multiply by

\frac{1 + sin ( x )}{1 + sin ( x )}

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

Expand

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

(1 + sin (x)) (1 - sin (x))

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (x)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (x)

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

Use the Pythagorean identity:

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

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

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

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply

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

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

Multiply fractions:

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

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

1 \cdot (1 + sin (x)) = 1 + sin (x)

1 \cdot (1 + sin (x))

Multiply:

1 \cdot (1 + sin (x)) = (1 + sin (x))

= (1 + sin (x))

Remove parentheses:

(a) = a

= 1 + sin (x)

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

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

Apply the fraction rule:

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

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

cos^{2} (x) cos (x) = cos^{3} (x)

cos^{2} (x) cos (x)

Apply exponent rule:

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

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

= cos^{2 + 1} (x)

Add the numbers:

2 + 1 = 3

= cos^{3} (x)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

(1 + cos (θ)) (csc (θ) - cot (θ)) = sin (θ)

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

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

2 - csc^{2} (z) = 1 - cot^{2} (z)

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

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