Prove each identity and state any restrictions

Identities should be true for all values of variables for which functions are defined. We may need to identify domain of the function and restrictions.

Example 7: Prove each identity and sate restrictions.
(a) \(\frac{1 - cos θ}{sin θ}\)

= \(\frac{sin θ}{1 + cos θ}\) (b) \(\frac{sec θ- 1}{sec θ + 1}\)

= (\(\frac{sin θ}{1 + cos θ}\))²

Solution 7: Prove each identity and sate restrictions

(a) \(\frac{1 - cos θ}{sin θ}\) = \(\frac{sin θ}{1 + cos θ}\)

Restrictions: denominator cannot be zero
cos ≠−1 ⇒ ≠180°+360
sin≠0⇒ ≠360°, ∈
Proof of Identity by Rationalization
Star from left hand side (LHS)

LHS = \(\frac{1 - cos θ}{sin θ}\)

= \(\frac{1 - cos θ}{sin θ} × \frac{1+ cos θ}{1+ cos θ}\)

= \(\frac{1- cos^2θ}{sinθ(1+ cos θ)}\)

= \(\frac{sin^2θ}{sin θ(1 + cos θ)}\)

= \(\frac{sinθ}{1 + cos θ}\) = RHS

(b)\(\frac{sec θ - 1}{sec θ+ 1}\) = (\(\frac{sin θ}{1 + cos θ}\))²

Restrictions: denominator cannot be zero
cos θ ≠−1 ⇒ ≠270°+360
cos θ ≠0 ⇒ ≠90°(2+1)
Proof of Identity by Rationalization

LHS = \(\frac{sec θ - 1}{ sec θ + 1}\) = \(\frac{\frac{1}{cos θ}}{\frac{1}{cos θ}} + 1\)
= \(\frac{1 - cos θ}{1 + cos θ}\)

= \(\frac{1 - cos θ}{1 + cos θ}× \frac{1+cos θ}{1+cos θ}\)

= \(\frac{1−cos^2}{(1+cos θ)^2}\)

= \(\frac{sin^2θ}{(1 + cos θ)^2}\) = (\(\frac{sin θ}{1 + cos θ}\))² = RHS