If sinA+tanA=p then sin2A?

sinA + tanA = p

sinA + sinA/cosA = p
times cosA
sinAcosA + sinA = pcosA
sinAcosA = pcosA - sinA

but sin 2A = 2sinAcosA
= 2pcosA - 2sinA

??

if x = sinA

tanA = x/√(1-x^2)
cosA = √(1-x^2)

x + x/√(1-x^2) = p
x^2/(1-x^2) = (p-x)^2

quartics don't solve simply. If you go to wolframalpha . com and type

solve x^2/(1-x^2) = (p-x)^2

you will see a horrendous solution in terms of p.

Anyway, pick a solution and evaluate

2x√(1-x^2)

tu bta sale

To find the value of sin2A, we can use the double-angle formula for sine:

sin2A = 2 * sinA * cosA

Let's find the values of sinA and cosA using the given equation sinA + tanA = p.

Since we know that tanA = sinA / cosA (tangent is the ratio of sine to cosine), we can substitute this into the equation:

sinA + sinA / cosA = p

To simplify the equation, we can multiply through by cosA:

(sinA * cosA) + sinA = p * cosA

Now, let's use the Pythagorean identity sin²A + cos²A = 1 to replace sin²A in the equation:

(1 - cos²A) + sinA = p * cosA

Rearranging the equation:

sinA + (1 - cos²A) = p * cosA

Expanding the equation:

sinA + 1 - cos²A = p * cosA

Rearranging again:

sinA = p * cosA - 1 + cos²A

Now, let's substitute sinA back into the equation for sin2A:

sin2A = 2 * sinA * cosA

sin2A = 2 * (p * cosA - 1 + cos²A) * cosA

simplify:

sin2A = 2p * cos²A - 2cosA + 2cos³A

Therefore, the value of sin2A is 2p * cos²A - 2cosA + 2cos³A.