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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Expand

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

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

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = 1, b = sin (x)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (x)

= 1 - sin^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

\frac{- cot ( x ) + 1}{tan ( x ) - 1} = cot (x)

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

4 cos^{2} (a) + 4 sin^{2} (a) = 4

\frac{sin ( a )}{tan ( a )} = cos (a)

4 sin^{2} (θ) + 8 cos (θ) = 3

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