Question

If (sec θ + tan θ) = p then show that `(sec θ - tan θ) = 1/p`. Hence, show that `cos θ = (2p)/(p^2 + 1)` and `sin θ = (p^2 - 1)/(p^2 + 1)`.

Answer 1

Given: If sec θ + tan θ = p.

Step-wise calculation:

1. Use the identity sec² θ – tan² θ = 1.

So (sec θ + tan θ)(sec θ – tan θ) = 1. 

Therefore `sec θ - tan θ = 1/(sec θ + tan θ) = 1/p`.

2. Add the two equations:

`(sec θ + tan θ) + (sec θ - tan θ) = p + 1/p`

⇒ `2 sec θ = p + 1/p` 

⇒ `sec θ = (p + 1/p)/2`

⇒ `sec θ = (p^2 + 1)/(2p)`

Hence `cos θ = 1/sec θ = (2p)/(p^2 + 1)`.

3. Subtract the second from the first:

`(sec θ + tan θ) - (sec θ - tan θ) = p - 1/p`

⇒ `2 tan θ = p - 1/p` 

⇒ `tan θ = (p - 1/p)/2`

⇒ `tan θ = (p^2 - 1)/(2p)`

4. Compute sin θ = tan θ · cos θ:

`sin θ = ((p^2 - 1)/(2p)) · ((2p)/(p^2 + 1))` 

= `(p^2 - 1)/(p^2 + 1)`