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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= cos^{2} (t)

= cos^{2} (t)

= cos^{2} (t)

Rewrite using trig identities

cos^{2} (t)

Use the Pythagorean identity:

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

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

= 1 - sin^{2} (t)

Factor

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

1 - sin^{2} (t)

Apply Difference of Two Squares Formula:

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

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

= (1 + sin (t)) (1 - sin (t))

= (1 + sin (t)) (1 - sin (t))

= (1 + sin (t)) (1 - sin (t))

We showed that the two sides could take the same form

\Rightarrow True

cos (2 0^{\circ}) = cos (θ) cos (θ) - sin (θ) sin (θ)

cos (θ) = \frac{4}{5}

\frac{2 sin ( 2 θ ) cos ( θ )}{2 sin ( 2 θ )} = cos (θ)

- sin (2 x) + cos (2 x) = 1

sin (x) (sec (x) tan (x)) = tan^{2} (x)

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