College Trigonometry, 2011a

750 Foundations of Trigonometry
2. Starting with the right hand side of tan(θ) =sin(θ) sec(θ), we use sec(θ) = 1
cos(θ)
1
sin(θ) sec(θ) =sin(θ)
cos(θ) = sin(θ)
cos(θ) = tan(θ),
where the last equality is courtesy of Theorem 10.6.
and find:
3. Expanding the left hand side of the equation gives: (sec(θ) − tan(θ))(sec(θ) +tan(θ)) =
sec 2 (θ) − tan 2 (θ). According to Theorem 10.8, sec 2 (θ) − tan 2 (θ) = 1. Putting it all together,
(sec(θ) − tan(θ))(sec(θ)+tan(θ)) = sec 2 (θ) − tan 2 (θ) =1.
4. While both sides of our last identity contain fractions, the left side affords us more opportunities
to use our identities. 5 Substituting sec(θ) = 1
sin(θ)
cos(θ)
and tan(θ) =
cos(θ) ,weget:
sec(θ)
1 − tan(θ)
=
=
1
cos(θ)
1 − sin(θ)
cos(θ)
( 1
cos(θ)
=
(
1 − sin(θ)
cos(θ)
1
cos(θ)
1 − sin(θ)
cos(θ)
)
(cos(θ))
)
(cos(θ))
1
=
cos(θ) − sin(θ) ,
which is exactly what we had set out to show.
· cos(θ)
cos(θ)
=
(1)(cos(θ)) −
1
( sin(θ)
cos(θ)
)
(cos(θ))
5. The right hand side of the equation seems to hold more promise. We get common denominators
and add:
3
1 − sin(θ) − 3
1+sin(θ)
=
3(1 + sin(θ))
(1 − sin(θ))(1 + sin(θ)) − 3(1 − sin(θ))
(1 + sin(θ))(1 − sin(θ))
= 3+3sin(θ)
1 − sin 2 (θ) − 3 − 3sin(θ)
1 − sin 2 (θ)
=
=
(3 + 3 sin(θ)) − (3 − 3sin(θ))
1 − sin 2 (θ)
6sin(θ)
1 − sin 2 (θ)
5 Or, to put to another way, earn more partial credit if this were an exam question!