Is (sec(θ)+tan(θ))(sec(θ)-tan(θ))=1 ?

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

prove (sec(θ)+tan(θ))(sec(θ)-tan(θ))=1

prove

(sec (θ) + tan (θ)) (sec (θ) - tan (θ)) = 1

True

Solution steps

(sec (θ) + tan (θ)) (sec (θ) - tan (θ)) = 1

Manipulating left side

(sec (θ) + tan (θ)) (sec (θ) - tan (θ))

Expand

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

(sec (θ) + tan (θ)) (sec (θ) - tan (θ))

Apply Difference of Two Squares Formula:

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

a = sec (θ), b = tan (θ)

= sec^{2} (θ) - tan^{2} (θ)

= sec^{2} (θ) - tan^{2} (θ)

Use the Pythagorean identity:

sec^{2} (θ) = tan^{2} (θ) + 1

sec^{2} (θ) - tan^{2} (θ) = 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

p ro v e \frac{cos ( u ) sec ( u )}{tan ( u )} = cot (u)

p ro v e sin (x) (1 + cot^{2} (x)) = csc (x)

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

p ro v e sec^{4} (x) - tan^{4} (x) = 1 + 2 tan^{2} (x)

p ro v e sin (- x) = - sin (x)

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