Is (cos^2(θ))/(1+sin(θ))+sin(θ)=1 ?

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

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prove (cos^2(θ))/(1+sin(θ))+sin(θ)=1

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

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

Cancel

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

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

Factor

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

1 - sin^{2} (θ)

Factor out common term

- 1

= - (sin^{2} (θ) - 1)

Factor

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

sin^{2} (θ) - 1

Rewrite

1

1^{2}

= sin^{2} (θ) - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (sin (θ) + 1) (sin (θ) - 1)

= - (sin (θ) + 1) (sin (θ) - 1)

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

Cancel the common factor:

sin (θ) + 1

= - (sin (θ) - 1)

Negate

- (sin (θ) - 1) = - sin (θ) + 1

= - sin (θ) + 1

= (- sin (θ) + 1) + sin (θ)

Remove parentheses:

(- a) = - a

= - sin (θ) + 1 + sin (θ)

Group like terms

= - sin (θ) + sin (θ) + 1

Add similar elements:

- sin (θ) + sin (θ) = 0

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

sin (2 t) = 2 sin (t) cos (t)

sin (2 a) + cos (a) = cos (a) (2 sin (a) + 1)

\frac{csc ( x )}{2 cos ( x )} = csc (2 x)

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

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

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