Help Prove this Trigonometric Identity :----------?

Question

Given :-
1. Sin ψ = {(3)^(1/2)} * Sin Φ
2. Tan Φ = x2/2a
3. Tan ψ= x1/a
4. x3*x2 = x1*2*a
5. (x3+2a)/(x1+x2) =[ {(x3)^2 +2*a*x3}^1/2]/a

Here Φ , ψ , x1 , x2 , x3 are variables
and 'a' is a constant

Prove that x1+x2 = {(2)^(1/2)}*a

Answer 1

I'll give you a start to the problem.

Starting with eqn 5,
(x3 + 2a)/(x1+x2) = sqrt[x3*(x3 + 2a)] / a
squaring both sides and rearranging
(x1+x2)^2 = a^2 * (x3 + 2a)^2 / [x3*(x3 + 2a)]
= a^2 * (x3 + 2a) / x3

From eqn 4, x3 = 2a*x1 / x2
Substituting in the above equation, we get
(x1+x2)^2 = a^2 * (x1 + x2) / x1
x1 + x2 = a^2 / x1 ...(eqn 6)

From eqn 1, sin^2 ψ = 3*sin^2 φ
also from eqns 2 and 3,
tan^2 φ = x2^2 / (4a^2)
tan^2 ψ = x1^2 / a^2
Using trig identities
sec^2 φ = (4a^2 + x2^2) / (4a^2)
sec^2 ψ = (a^2 + x1^2) / a^2
Using tan^2 ψ = sin^2 ψ * sec^2 ψ
we get
x1^2 / a^2 = 3*sin^2 φ * (a^2 + x1^2) / a^2
rearranging,
a^2 * (4*x1^2 - 3*x2^2) = 2 * x1^2 * x2^2 ... (eqn 7)

Now we need to solve eqns 6 and 7 simultaneously.
At this point it will greatly simplify things if we let
y1 = x1/a, y2 = x2/a
So y1 + y2 = 1/y1 ...(eqn 6a)
4y1^2 - 3y2^2 = 2y1^2*y2^2 ... eqn(7a)