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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

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

1 - sin^{2} (a)

Factor out common term

- 1

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

Factor

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

sin^{2} (a) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

= (sin (a) + 1) (sin (a) - 1)

= - (sin (a) + 1) (sin (a) - 1)

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

Cancel

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

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

- sin (a) + 1 = - (sin (a) - 1)

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

Refine

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

Cancel the common factor:

sin (a) - 1

= sin (a) + 1

= sin (a) + 1

= sin (a) + 1

= sin (a) + 1

= 1 + sin (a)

We showed that the two sides could take the same form

\Rightarrow True

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

2 tan^{2} (x) = \frac{2 sin ^{2} ( x )}{cos ^{2} ( x )}

csc (θ) \cdot tan (θ) = sec (θ)

cot (θ) \cdot csc (θ) \cdot sec (θ) = csc^{2} (θ)

sin (x) + cos (x) + cot (x) = csc (x)

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