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.