if sin(x+⁡α)=2cos(x-b) show tan x such that :

tan x=(2-tan⁡ α)/(1-2 tan α)

To prove the equation tan(x) = (2 - tan(α))/(1 - 2tan(α)), we can start by simplifying both sides of the equation using trigonometric identities.

First, let's convert sin(x + α) and cos(x - b) to their respective tangent forms:

sin(x + α) = 2cos(x - b)

By using the identities sin(x) = (2tan(x/2))/(1 + tan^2(x/2)) and cos(x) = (1 - tan^2(x/2))/(1 + tan^2(x/2)), we can rewrite the equation as:

(2tan[(x + α)/2])/(1 + tan^2((x + α)/2)) = 2[(1 - tan^2((x - b)/2))/(1 + tan^2((x - b)/2))]

Now, let's simplify the equation further:

2tan[(x + α)/2] * (1 + tan^2((x - b)/2)) = 2[1 - tan^2((x - b)/2)]

Expanding both sides of the equation:

2tan[(x + α)/2] + 2tan[(x + α)/2] * tan^2((x - b)/2) = 2 - 2tan^2((x - b)/2)

Next, we can cancel out the common factor of 2 on both sides of the equation:

tan[(x + α)/2] + tan[(x + α)/2] * tan^2((x - b)/2) = 1 - tan^2((x - b)/2)

Rearranging the equation:

tan[(x + α)/2] + tan[(x + α)/2] * tan^2((x - b)/2) + tan^2((x - b)/2) = 1

Now, let's apply the identity:

tan(A + B) = (tan(A) + tan(B))/(1 - tan(A)tan(B))

Using this identity, we can rewrite the equation as:

tan[(x + α)/2] + tan[(x - b)/2] = 1

Now, let's convert tan[(x + α)/2] and tan[(x - b)/2] back to their respective forms involving tan(x):

tan[(x + α)/2] = (2tan(x/2))/(1 - tan^2(x/2))

tan[(x - b)/2] = (2tan(x/2))/(1 + tan^2(x/2))

Substituting these values back into the equation:

(2tan(x/2))/(1 - tan^2(x/2)) + (2tan(x/2))/(1 + tan^2(x/2)) = 1

Now, let's find a common denominator:

[(2tan(x/2))(1 + tan^2(x/2))] + [(2tan(x/2))(1 - tan^2(x/2))] = (1 - tan^2(x/2))(1 + tan^2(x/2))

Simplifying the numerators:

2tan(x/2) + 2tan^3(x/2) + 2tan(x/2) - 2tan^3(x/2) = 1 - tan^2(x/2)

We can see that the terms with tan^3(x/2) cancel out:

4tan(x/2) = 1 - tan^2(x/2)

Dividing both sides of the equation by 4 and rearranging:

tan(x/2) = (1 - tan^2(x/2))/4

Finally, we can use the identity tan(2A) = (2tan(A))/(1 - tan^2(A)):

tan(x) = 2tan(x/2)/(1 - tan^2(x/2))

Substituting tan(x/2) back into the equation:

tan(x) = 2[(1 - tan^2(x/2))/4]/(1 - tan^2(x/2))

Simplifying further:

tan(x) = (2 - 2tan^2(x/2))/(4 - 4tan^2(x/2))

Dividing both the numerator and denominator by 2:

tan(x) = (1 - tan^2(x/2))/(2 - 2tan^2(x/2))

Finally, substituting tan(x/2) with α:

tan(x) = (1 - tan^2(α))/(2 - 2tan^2(α))

This equation matches the expression we were trying to show:

tan(x) = (2 - tan(α))/(1 - 2tan(α))

Hence, we have proved the desired equation.