if tanX=a/b, then show that asinX-bcosX/asinX+bcosx = a^2-b^2/a^2+b^2
To prove the given equation, we'll start by substituting the value of tan(X) into the expression:
tan(X) = a/b
We can rewrite this as:
sin(X)/cos(X) = a/b
Now, let's manipulate the expression to obtain the desired result. Multiply both sides of the equation by b*cos(X):
b*cos(X)*sin(X)/cos(X) = (a/b)*(b*cos(X))
The cos(X) terms cancel out on the left side:
b*sin(X) = a*cos(X)
Now, let's square both sides of the equation to get rid of the square root:
(b*sin(X))^2 = (a*cos(X))^2
Expanding both sides:
b^2*sin^2(X) = a^2*cos^2(X)
Next, let's use the identity sin^2(X) + cos^2(X) = 1:
b^2*(1 - cos^2(X)) = a^2*cos^2(X)
Distributing the b^2 on the left side:
b^2 - b^2*cos^2(X) = a^2*cos^2(X)
Rearranging the terms:
b^2 = (a^2 + b^2)*cos^2(X)
Dividing both sides by (a^2 + b^2):
b^2/(a^2 + b^2) = cos^2(X)
Now, let's use the identity sin^2(X) = 1 - cos^2(X):
sin^2(X) = 1 - b^2/(a^2 + b^2)
Multiplying through by (a^2 + b^2):
(a^2 + b^2)*sin^2(X) = a^2 + b^2 - b^2
Simplifying:
(a^2 + b^2)*sin^2(X) = a^2
Now, divide both sides by (a^2 + b^2):
sin^2(X) = a^2/(a^2 + b^2)
Finally, taking the square root of both sides:
sin(X) = a/sqrt(a^2 + b^2)
Now we have the value of sin(X), we can substitute this into the given equation:
(a*sin(X) - b*cos(X))/(a*sin(X) + b*cos(X))
Substituting sin(X) = a/sqrt(a^2 + b^2) and rearranging:
(a*(a/sqrt(a^2 + b^2)) - b*cos(X))/(a*(a/sqrt(a^2 + b^2)) + b*cos(X))
Simplifying:
(a^2 - b*cos(X)*sqrt(a^2 + b^2))/(a^2 + b*cos(X)*sqrt(a^2 + b^2))
Now, let's simplify the expression by multiplying both the numerator and denominator by sqrt(a^2 + b^2):
((a^2 - b*cos(X)*sqrt(a^2 + b^2))/(a^2 + b*cos(X)*sqrt(a^2 + b^2)))*(sqrt(a^2 + b^2)/sqrt(a^2 + b^2))
This results in:
(a^2*sqrt(a^2 + b^2) - b*cos(X)*(a^2 + b^2))/(a^2*(sqrt(a^2 + b^2)) + b*cos(X)*(sqrt(a^2 + b^2)))
Expanding the terms:
a^2*sqrt(a^2 + b^2) - b*cos(X)*a^2 - b*cos(X)*b^2)/(a^2*sqrt(a^2 + b^2) + b*cos(X)*a^2 + b*cos(X)*b^2)
Now we can cancel out common factors:
(a^2*sqrt(a^2 + b^2) - b*a^2 - b*b^2*cos(X))/(a^2*sqrt(a^2 + b^2) + b*a^2 + b*b^2*cos(X))
Simplifying the expression further:
(a^2*sqrt(a^2 + b^2) - a^2*b - b^3*cos(X))/(a^2*sqrt(a^2 + b^2) + a^2*b + b^3*cos(X))
Now, let's notice that both the numerator and the denominator have a common factor of a^2+b^2:
(a^2+b^2)*(sqrt(a^2 + b^2) - b*cos(X))/(a^2+b^2)*(sqrt(a^2 + b^2) + b*cos(X))
Canceling out the common factors:
(sqrt(a^2 + b^2) - b*cos(X))/(sqrt(a^2 + b^2) + b*cos(X))
And this is equal to:
(a^2 - b^2)/(a^2 + b^2)
Which matches the desired result. Hence, the equation (asinX - bcosX) / (asinX + bcosX) = (a^2 - b^2) / (a^2 + b^2) is proven true.