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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

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

1 - sin^{2} (x)

Factor out common term

- 1

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

Factor

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

sin^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

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

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

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

Cancel

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

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

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

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

Refine

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

Cancel the common factor:

sin (x) - 1

= sin (x) + 1

= sin (x) + 1

= sin (x) + 1

= sin (x) + 1

= 1 + sin (x)

We showed that the two sides could take the same form

\Rightarrow True

sec (x) cot (x) = csc (x)

tan (2 x) = \frac{2}{cot ( x ) - tan ( x )}

\frac{sin ( 2 x )}{sin ( x )} - \frac{cos ( 2 x )}{cos ( x )} = sec (x)

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

(csc^{2} (x) - 1) sec (x) = cos (x) csc^{2} (x)

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