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

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

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

prove

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

True

Solution steps

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

Manipulating right side

\frac{1}{sec ^{2} ( x )}

Express with sin, cos

\frac{1}{sec ^{2} ( x )}

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

= \frac{1}{( \frac{1}{c o s ( x )} ) ^{2}}

Simplify

\frac{1}{( \frac{1}{c o s ( x )} ) ^{2}} : cos^{2} (x)

\frac{1}{( \frac{1}{c o s ( x )} ) ^{2}}

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= \frac{1}{\frac{1}{c o s ^{2} ( x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

Apply rule

\frac{a}{1} = a

= cos^{2} (x)

= cos^{2} (x)

= cos^{2} (x)

Rewrite using trig identities

cos^{2} (x)

Use the Pythagorean identity:

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

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

= 1 - sin^{2} (x)

Factor

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

1 - sin^{2} (x)

Apply Difference of Two Squares Formula:

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

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

sin (3 x) cos (2 x) - cos (3 x) sin (2 x) = sin (x)

4 cot^{2} (y) (sec^{2} (y) - 1) = 4

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

sin^{2} (x) sec (x) = \frac{sin ( x )}{cot ( x )}

cot^{2} (x) (1 + tan^{2} (x)) = csc^{2} (x)

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