Is 1-((cos^2(θ)))/(1+sin(θ))=sin(θ) ?

The answer to whether 1-((cos^2(θ)))/(1+sin(θ))=sin(θ) is True

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prove 1-((cos^2(θ)))/(1+sin(θ))=sin(θ)

prove

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

True

Solution steps

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

Manipulating left side

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Rewrite using trig identities

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

Simplify

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

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

Factor

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

sin^{2} (θ) + sin (θ)

Apply exponent rule:

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

sin^{2} (θ) = sin (θ) sin (θ)

= sin (θ) sin (θ) + sin (θ)

Factor out common term

sin (θ)

= sin (θ) (sin (θ) + 1)

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

Cancel the common factor:

sin (θ) + 1

= sin (θ)

= sin (θ)

= sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Is 1-((cos^2(θ)))/(1+sin(θ))=sin(θ) ?
The answer to whether 1-((cos^2(θ)))/(1+sin(θ))=sin(θ) is True