Is 1-(sin^2(θ))/(1+cos(θ))=cos(θ) ?

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

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prove 1-(sin^2(θ))/(1+cos(θ))=cos(θ)

prove

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

True

Solution steps

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

Manipulating left side

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

Simplify

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

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

Convert element to fraction:

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

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

1 \cdot (1 + cos (θ)) = 1 + cos (θ)

1 \cdot (1 + cos (θ))

Multiply:

1 \cdot (1 + cos (θ)) = (1 + cos (θ))

= (1 + cos (θ))

Remove parentheses:

(a) = a

= 1 + cos (θ)

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

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

cos (θ) + cos^{2} (θ)

Apply exponent rule:

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

cos^{2} (θ) = cos (θ) cos (θ)

= cos (θ) + cos (θ) cos (θ)

Factor out common term

cos (θ)

= cos (θ) (1 + cos (θ))

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

Cancel the common factor:

1 + cos (θ)

= cos (θ)

= cos (θ)

= cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{cos ( 2 X )}{sin ( X )} + \frac{sin ( 2 X )}{cos ( X )} = csc (X)

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

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

tan^{5} (x) = tan^{3} (x) sec^{2} (x) - tan^{3} (x)

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