How do you solve the identity (sec(t)+1)(sec(t)-1)=tan^2(t)(sec(t)+1)(sec(t)1)=tan2(t)?

1 Answer
Sep 15, 2015

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

Explanation:

Since cos^2(t)+sin^2(t)=1cos2(t)+sin2(t)=1 for all tt (this is the Pythagorean identity), it follows that 1+(sin^2(t))/(cos^2(t))=1/(cos^2(t))1+sin2(t)cos2(t)=1cos2(t) for all tt for which it is defined.

By definition, tan(t)=(sin(t))/(cos(t))tan(t)=sin(t)cos(t) and sec(t)=1/(cos(t))sec(t)=1cos(t). Therefore, 1+tan^2(t)=sec^2(t)1+tan2(t)=sec2(t) for all tt for which it is defined.

Rearranging this equation leads to sec^2(t)-1=tan^2(t)sec2(t)1=tan2(t) and then factoring leads us to conclude that the desired equation is an identity: (sec(t)+1)(sec(t)-1)=tan^2(t)(sec(t)+1)(sec(t)1)=tan2(t) for all tt for which it is defined.