If sec θ + tan θ = p, prove that
(i)
(ii)
(iii)
(i) Given: sec θ + tan θ = p ……(1)
Then, (sec θ + tan θ) × = p
⇒ = p
⇒ = p
⇒ sec θ – tan θ = (1/p) ……(2)
Adding equation (1) and (2), we get:
2 sec θ = p + (1/p)
⇒ sec θ =
Therefore, sec θ =
(ii) Given: sec θ + tan θ = p ……(1)
Then, (sec θ + tan θ) × = p
⇒ = p
⇒ = p
⇒ sec θ – tan θ = (1/p) ……(2)
Subtracting equation (2) from (1), we get:
2tan θ = p – (1/p)
⇒ tan θ =
(iii) Since sin θ = tan θ/sec θ
=
=
=