Trigo question URGENT?

Question

Given tanA= (√(1-p^2))/p

find tan(A+pi/4)

not enough

Answer 1

Use the sum formula:

tan(A + pi/4) = (tan(A) + tan(pi/4))/(tan(A) - tan(pi/4))

tan(pi/4) = SQRT(2)/2

tanA= (SQRT(1 - p^2))/p

tan(A + pi/4) = [(SQRT(1 - p^2))/p + SQRT(2)/2]/[(SQRT(1 - p^2))/p - SQRT(2)/2]

Is this good enough?

Answer 2

since π/4 = 45°, then

tan(A + 45°) = [(√1-p²)/p] + 1 all over 1-[(√1-p²)/p](1)

=p + (√1-p²) * 1/[p - (√1-p²)]

then

p + (√1-p²) = p - (√1-p²) thus cancelling p, leaving the radicals.

(√1-p²)² = -1 (√1-p²)² thereby removing the radical sign...

1 - p²= -1 + p²
leaving

p= ±1.

Then solve for A.p={}

tan(A + 45°) = ±1(√1-{±1}²) + 1 all over 1 - (±1)(1)(√1-p²).

yielding TWO solutions...

the positive root of A=1 + √2 all over 1 - √2

which yields A=-5.828427124

and the negative root = -1 + √2 all over 1 + √2

which yields A=0.171572875.

Quite the way I understand it.

Answer 3

tan(A+pi/4)=[tanA+tan(pi/4)]/[1-tanAtan(pi/4)]
=[tanA+1]/[1-tanA]=[1+tanA]/[1-tanA].
Substitute for tanA and simplify.

Answer 4

tan(A+B)=(tanA+tanB)/(1-tanAtanB)
Since tan(pi/4)=1
tan(A+pi/4)=((sqrt(1-p^2)/p+1)/(1-sqrt(1-p^2)/p)
=((sqrt(1-p^2)+p)/(p-sqrt(1-p^2))