Question

If sec θ + tan θ = p, prove that: sin θ = `(p^2 - 1)/(p^2 + 1)`

Answer 1

sec² θ − tan² θ = 1

(sec θ + tan θ) (sec θ - tan θ) = 1

p(sec θ - tan θ) = 1

sec θ - tan θ = `1/p`

2 sec θ = `p + 1/p`

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

sec θ = `(p^2 + 1)/(2p)`

2 tan θ = `p - 1/p`

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

tan θ = `(p^2 - 1)/(2p)`

We know that tan θ = `sin θ/cos θ`, which can also be written as:

sin θ = `tan θ/sec θ`

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

sin θ = `(p^2 - 1)/(p^2 + 1)`