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

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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^{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

\frac{sin ^{2} ( x )}{cos ^{2} ( x )} = tan^{2} (x)

sin (x + y) = sin (x) cos (y) + sin (y) cos (x)

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

sin (u) csc (u) - cos^{2} (u) = sin^{2} (u)

sin (x) + cot (x) cos (x) = csc (x)

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