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

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

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prove (sec(t)-1)(sec(t)+1)=tan^2(t)

prove

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

True

Solution steps

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

Manipulating left side

(sec (t) - 1) (sec (t) + 1)

Expand

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

(sec (t) + 1) (sec (t) - 1)

Apply Difference of Two Squares Formula:

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

a = sec (t), b = 1

= sec^{2} (t) - 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= sec^{2} (t) - 1

= sec^{2} (t) - 1

Use the Pythagorean identity:

sec^{2} (t) = tan^{2} (t) + 1

sec^{2} (t) - 1 = tan^{2} (t)

= tan^{2} (t)

We showed that the two sides could take the same form

\Rightarrow True

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cos^{2} (a) cot^{2} (a) = cot^{2} (a) - cos^{2} (a)

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