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

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

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

prove

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

True

Solution steps

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

Manipulating right side

\frac{1 - sin ( θ )}{cos ^{2} ( θ )}

Rewrite using trig identities

\frac{1 - sin ( θ )}{cos ^{2} ( θ )}

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Cancel

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

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

sin (θ) - 1 = - (- sin (θ) + 1)

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

Refine

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

Cancel the common factor:

1 - sin (θ)

= \frac{1}{sin ( θ ) + 1}

= \frac{1}{sin ( θ ) + 1}

= \frac{1}{sin ( θ ) + 1}

= \frac{1}{sin ( θ ) + 1}

We showed that the two sides could take the same form

\Rightarrow True

sin (x) cos (x) tan (x) cot (x) = 1

(1 - cos (x)) = cos (x) - 1

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

sin (θ - \frac{π}{2}) + sin (θ + \frac{π}{2}) = 0

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

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