1. Solution

2. sinθ cosθ
csc θ sec θ
2=
2
tan θ + cot θ sin2θ ­ cos2θ
Change everything to sin and
cos so it is easier to work with.
1
1
1 1
sinθ cosθ cscθ = secθ =
cosθ
sinθ
2
cos θ cos2θ
2
sin θ sin2θ
2
+ tan2θ = 2
cot θ =
2
2 cos θ
2 sin θ
cos θ sin θ

3. sinθ cosθ
csc θ sec θ
2=
2
tan θ + cot θ sin2θ ­ cos2θ
1
Find the Lowest Common
sinθ cosθ Denominator
() ()
cos2θ cos2θ sin2θ
2
sin θ
+2
4
4
sin θ + cos θ cos2θ sin2θ cos θ sin2θ
2
2
sin θ cos θ

4. sinθ cosθ
csc θ sec θ
2=
2 2 2
tan θ + cot θ sin θ ­ cos θ
1
Find the Difference of
sinθ cosθ
Squares of
4
4
sin θ + cos θ
2
(sin2θ ­ cos2θ) (sin2θ + cos θ)
2
2
sin θ cos θ

5. sinθ cosθ
csc θ sec θ
2=
2
tan θ + cot θ sin2θ ­ cos2θ
1
sinθ cosθ We know this identity
2
2
sin θ + cos θ = 1
2
(1) (sin2θ ­ cos θ)
cos2θ
2
sin θ

6. sinθ cosθ
csc θ sec θ
2=
2
tan θ + cot θ sin2θ ­ cos2θ
When we divide by a
sin2θ cos2θ
1
number we can
(sin2θ ­ cos2θ)
sinθ cosθ multiply by the
reciprocal

7. sinθ cosθ
csc θ sec θ
2=
2 2 2
tan θ + cot θ sin θ ­ cos θ
sin2θ cos2θ Simplify and
1
there's the
(sin2θ ­ cos2θ)
sinθ cosθ
answer
sinθ cosθ
Q.E.D.
2 2
sin θ ­ cos θ