Introduction
To show that (P² - 1) / (P² + 1) = sin θ, given that sec θ + tan θ = P, we start from the trigonometric identities of secant and tangent.
Step 1: Express P in terms of sin and cos
- Recall that:
- sec θ = 1/cos θ
- tan θ = sin θ/cos θ
- Therefore, we can rewrite P as:
- P = 1/cos θ + sin θ/cos θ
- P = (1 + sin θ) / cos θ
Step 2: Square P
- Now, square both sides:
- P² = [(1 + sin θ) / cos θ]²
- P² = (1 + 2sin θ + sin² θ) / cos² θ
- P² = (1 + 2sin θ + sin² θ) / (1 - sin² θ) (using cos² θ = 1 - sin² θ)
Step 3: Simplifying P² - 1 and P² + 1
- Calculate P² - 1:
- P² - 1 = [(1 + 2sin θ + sin² θ) - (1 - sin² θ)] / (1 - sin² θ)
- = (2sin θ + 2sin² θ) / (1 - sin² θ)
- = 2sin θ(1 + sin θ) / (1 - sin² θ)
- Calculate P² + 1:
- P² + 1 = [(1 + 2sin θ + sin² θ) + (1 - sin² θ)] / (1 - sin² θ)
- = (2 + 2sin θ) / (1 - sin² θ)
- = 2(1 + sin θ) / (1 - sin² θ)
Step 4: Form the ratio
- Now form the ratio (P² - 1) / (P² + 1):
- (P² - 1) / (P² + 1) = [2sin θ(1 + sin θ)] / [2(1 + sin θ)]
- Canceling out (1 + sin θ):
- = sin θ
Conclusion
Thus, we have shown that (P² - 1) / (P² + 1) = sin θ, as required. This derivation employs fundamental trigonometric identities and algebraic manipulation to arrive at the conclusion.