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

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

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

prove

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

True

Solution steps

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

Manipulating right side

\frac{1}{sec ^{2} ( θ )}

Express with sin, cos

\frac{1}{sec ^{2} ( θ )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

(\frac{1}{cos ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

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

Apply the fraction rule:

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

= \frac{cos ^{2} ( θ )}{1}

Apply rule

\frac{a}{1} = a

= cos^{2} (θ)

= cos^{2} (θ)

= cos^{2} (θ)

Rewrite using trig identities

cos^{2} (θ)

Use the Pythagorean identity:

cos^{2} (θ) + sin^{2} (θ) = 1

cos^{2} (θ) = 1 - sin^{2} (θ)

= 1 - sin^{2} (θ)

Factor

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

1 - sin^{2} (θ)

Apply Difference of Two Squares Formula:

θ^{2} - y^{2} = (θ + y) (θ - y)

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

= (1 + sin (θ)) (1 - sin (θ))

= (1 + sin (θ)) (1 - sin (θ))

= (1 + sin (θ)) (1 - sin (θ))

= (1 - sin (θ)) (1 + sin (θ))

We showed that the two sides could take the same form

\Rightarrow True

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

sin^{2} (x) - 2 cos^{2} (x) = 1 - 3 cos^{2} (x)

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

\frac{sin ( x )}{1 + cos ( x )} + cot (x) = csc (x)

cos (θ - \frac{π}{2}) = sin (θ)

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