Question

If seco+tan =p, then what is the value of seco-tan ?​

Answer 1

Answer:

If sec θ+tan θ=p, then how do I show that sin θ=(p^2 - 1)/(p^2 + 1)?

Given that sec A + tan A = p.

Square both sides to get

sec^2 A +2 sec A tan A + tan^2 A = p^2

(1 + tan^2 A) +2 sec A tan A + tan^2 A = p^2, or

tan^2 A +2 sec A tan A + tan^2 A = p^2 - 1, or

2tan^2 A +2 sec A tan A = p^2 - 1, or

2 tan A (tan A + sec A) = p^2 - 1, or

2 tan A*p = p^2 - 1, or

tan A = (p^2 - 1)/2p

Consider a right angled triangle whose altitude is (p^2 - 1) and the base is 2p. The the hypotenuse = [(p^2 - 1)^2 + (2p)^2]^0.5

= [p^4–2p^2+1+4p^2]^0.5

= [p^4+2p^2+1]^0.5

= (p^2+1)

Hence sin A = altitude/hypotenuse = (p^2-1)/(p^2+1).

Proved.