Question

(sinA +tgA/tgA)^2 - 2cos A = 1 +cos^2A

Answer 1

Решение во вложении.

Answer 2

Let's simplify the left-hand side of the equation first:

(sinA + tgA/tgA)^2 = (sinA + 1)^2 = sin^2A + 2sinA + 1 (using the identity tgA = sinA/cosA and simplifying)

So the original equation becomes:

sin^2A + 2sinA + 1 - 2cosA = 1 + cos^2A

Rearranging and using the identity sin^2A + cos^2A = 1, we get:

sin^2A - 2cosA + cos^2A + 2sinA = 0

(sinA - cosA)^2 + 2sinA - 2cosA = 0

Now we can use the identity sinA = cos(π/2 - A) and simplify further:

(cos(π/2 - A) - cosA)^2 + 2cos(π/2 - A) - 2cosA = 0

(cosπ/2 - A)cosA - sin(π/2 - A)sinA + 2sinπ/2 - A - 2cosA = 0

sinA - cosA - 2sinA - 2cosA = 0

-3sinA - 3cosA = 0

Dividing both sides by -3 gives:

sinA + cosA = 0

Using the identity sinA = cos(π/2 - A), we get:

cos(π/2 - A) + cosA = 0

2cos(π/4)cos(π/4 - A) = 0

cos(π/4 - A) = 0

π/4 - A = π/2 + kπ (where k is any integer)

A = -π/4 + kπ

So the solutions to the equation are:

A = -π/4 + kπ (where k is any integer)