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How do you solve the identity #(sec(t)+1)(sec(t)-1)=tan^2(t)#?

Answer 1

You can prove this identity by using the definition of secant and tangent and by using the Pythagorean identity.

Since #cos^2(t)+sin^2(t)=1# for all #t# (this is the Pythagorean identity), it follows that #1+(sin^2(t))/(cos^2(t))=1/(cos^2(t))# for all #t# for which it is defined.

By definition, #tan(t)=(sin(t))/(cos(t))# and #sec(t)=1/(cos(t))#. Therefore, #1+tan^2(t)=sec^2(t)# for all #t# for which it is defined.

Rearranging this equation leads to #sec^2(t)-1=tan^2(t)# and then factoring leads us to conclude that the desired equation is an identity: #(sec(t)+1)(sec(t)-1)=tan^2(t)# for all #t# for which it is defined.

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Answer 2

Grace Ashcraft

To solve the identity ((\sec(t) + 1)(\sec(t) - 1) = \tan^2(t)), expand the left side and simplify:

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

This follows from the trigonometric identity (\sec^2(t) - 1 = \tan^2(t)). So, the identity ((\sec(t) + 1)(\sec(t) - 1) = \tan^2(t)) is proven.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.