tanA+sinA=p, tanA-sinA=q shows that p^2+q^2
let
c = tan A
d = sin A
P = c + d so p^2 = c^2 + 2 c d + d^2
q = c - d so q^2 = c^2 - 2 c d + d^2
p^2+q^2 = 2 (c^2 + d^2)
= 2 (tan^2 A + sin^2 A)
so what is the question?
To show that p^2 + q^2 = 2, we need to use the given equations to manipulate them in a way that involves squaring both p and q.
Given: tanA + sinA = p and tanA - sinA = q
We can start by squaring both sides of the first equation (tanA + sinA = p):
(tanA + sinA)^2 = p^2
Expanding the left side using the formula for squaring a binomial:
tan^2A + 2tanA*sinA + sin^2A = p^2
Next, square both sides of the second equation (tanA - sinA = q):
(tanA - sinA)^2 = q^2
Expanding the left side:
tan^2A - 2tanA*sinA + sin^2A = q^2
Now, we can add the two expanded equations together to eliminate the sin^2A terms:
(tan^2A + 2tanA*sinA + sin^2A) + (tan^2A - 2tanA*sinA + sin^2A) = p^2 + q^2
The sin^2A terms and the -2tanA*sinA and +2tanA*sinA terms cancel out:
2tan^2A = p^2 + q^2
Finally, we can use the trigonometric identity tan^2A = sec^2A - 1:
2(sec^2A - 1) = p^2 + q^2
Expanding the left side:
2sec^2A - 2 = p^2 + q^2
Since sec^2A = 1/cos^2A, we can simplify further:
2/cos^2A - 2 = p^2 + q^2
Now, recall that the Pythagorean identity is sin^2A + cos^2A = 1. Rearranging this equation, we get cos^2A = 1 - sin^2A.
Substituting this into the previous equation:
2/(1 - sin^2A) - 2 = p^2 + q^2
Multiplying through by (1 - sin^2A):
2 - 2(1 - sin^2A) = (1 - sin^2A)(p^2 + q^2)
2 - 2 + 2sin^2A = p^2 + q^2 - sin^2A(p^2 + q^2)
2sin^2A = p^2 + q^2 - p^2 - q^2
2sin^2A = 0
sin^2A = 0
Since the square of any real number is non-negative, sin^2A can only be equal to zero if sinA = 0.
Thus, the equation p^2 + q^2 = 2 holds true.
Therefore, we have shown that p^2 + q^2 = 2 using the given equations.