trigonometric identity

Question

If tan A + sec A =L , then prove that (L^2+L)/2L = sec A

Answer 1

We have formula (sec^2 A - tan^2 A) = 1

From formula (a^2 - b^2) = (a + b) (a-b)

We can write (sec^2 A - tan^2 A) = (sec A + tan A) (sec A - tan A)

(sec^ A - tan^2 A) = 1

(sec A + tan A)(sec A - tan A) = 1

So we can write (sec A - tan A) = 1/(sec A + tan A)

We have (sec A  + tan A) = L

so (sec A - tan A)  = 1/L

From given question (tan A + sec A) = L

Squaring on both sides (tan A +sec A)^2 = L^2

From formula (a + b)^2 = (a^2 + b^2 + 2ab)

(tan^2 A) + (sec^2 A) +(2(tan A)(sec A) = L^2  

[From formula (sec^2 A - tan^2 A) = 1     We can write (tan^ 2A) = (sec^2 A -1)]

(sec^2 A - 1 +sec^2 A + 2sec Atan A) = L^2  
[from question (tan A + sec A) = L   (tan A) = (L - sec A)]   

(2sec^2 A - 1 +2sec A(L - sec A) = L^2

(2sec^2 A - 1 +2sec A(L) - 2sec^2 A = L^2                                                   

By cancelling +2sec^2 A and -2sec^ A

We get  2sec A(L) - 1 = L^2

2sec A(L) = L^2 + 1

sec A = L^2 + 1/2L

So finally it was proved that (sec A) = L^2 + L)/2L