Is (sec(θ)-1)(sec(θ)+1)=tan^2(θ) ?

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

prove (sec(θ)-1)(sec(θ)+1)=tan^2(θ)

prove

(sec (θ) - 1) (sec (θ) + 1) = tan^{2} (θ)

True

Solution steps

(sec (θ) - 1) (sec (θ) + 1) = tan^{2} (θ)

Manipulating left side

(sec (θ) - 1) (sec (θ) + 1)

Expand

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

(sec (θ) + 1) (sec (θ) - 1)

Apply Difference of Two Squares Formula:

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

a = sec (θ), b = 1

= sec^{2} (θ) - 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= sec^{2} (θ) - 1

= sec^{2} (θ) - 1

Use the Pythagorean identity:

sec^{2} (θ) = tan^{2} (θ) + 1

sec^{2} (θ) - 1 = tan^{2} (θ)

= tan^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e 1 + tan^{2} (x) = sec^{2} (x)

p ro v e sin^{3} (2 x) = (\frac{1}{2} sin (2 x)) (1 - cos (4 x))

p ro v e sin (x + y) + sin (x - y) = 2 sin (x) cos (y)

p ro v e cos (2 θ) = cos^{2} (θ) - sin^{2} (θ)

p ro v e sin (3 θ) = 3 sin (θ) - 4 sin^{3} (θ)

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