ifasinx=bcosx=2ctanx/1-tan^2 x prove that(a^2-b^2)^2 =4c^2 (a^2+b^2
ifasinx=bcosx=2ctanx/1-tan^2 x prove that(a^2-b^2)^2 =4c^2 (a^2+b^2
To prove the equation (a^2 - b^2)^2 = 4c^2(a^2 + b^2), we'll start by manipulating the given trigonometric expressions:
Since a*sin(x) = b*cos(x), we can divide both sides of the equation by b*sin(x):
a/b = cos(x)/sin(x)
Knowing that cos(x)/sin(x) = cot(x) (the reciprocal of tan(x)), we can rewrite the equation as:
a/b = cot(x)
Squaring both sides of the equation:
(a/b)^2 = cot^2(x)
Now, let's manipulate the other trigonometric expression:
b*cos(x) = 2c*tan(x)/(1 - tan^2(x))
Dividing both sides by b and simplifying:
cos(x)/b = 2c*tan(x)/(b*(1 - tan^2(x)))
Since 1 - tan^2(x) = sec^2(x), we can rewrite the equation as:
cos(x)/b = 2c*tan(x)/(b*sec^2(x))
Rewriting tan(x) as sin(x)/cos(x):
cos(x)/b = 2c*(sin(x)/cos(x))/(b*sec^2(x))
Simplifying further:
cos(x)/b = 2c*sin(x)/(b*cos(x))
Dividing both sides by cos(x)/b:
1 = 2c*sin(x)/(b*cos(x))
Simplifying again:
2c*sin(x) = b*cos(x)
Squaring both sides:
(2c*sin(x))^2 = (b*cos(x))^2
4c^2*sin^2(x) = b^2*cos^2(x)
Using the trigonometric identity sin^2(x) + cos^2(x) = 1:
4c^2*(1 - cos^2(x)) = b^2*cos^2(x)
Expanding:
4c^2 - 4c^2*cos^2(x) = b^2*cos^2(x)
Rearranging:
4c^2 = b^2*cos^2(x) + 4c^2*cos^2(x)
4c^2 = (b^2 + 4c^2)*cos^2(x)
Since cos^2(x) = 1 - sin^2(x), we can substitute:
4c^2 = (b^2 + 4c^2)*(1 - sin^2(x))
Expanding again:
4c^2 = b^2 + 4c^2 - b^2*sin^2(x) - 4c^2*sin^2(x)
Rearranging:
0 = b^2 - b^2*sin^2(x) - 4c^2*sin^2(x)
Factoring out common terms:
0 = b^2*(1 - sin^2(x)) - 4c^2*sin^2(x)
Using the trigonometric identity sin^2(x) + cos^2(x) = 1 again:
0 = b^2*cos^2(x) - 4c^2*sin^2(x)
We know from our earlier manipulation that:
4c^2*sin^2(x) = b^2*cos^2(x)
Substituting this back into the equation:
0 = b^2*cos^2(x) - b^2*cos^2(x)
0 = 0
The equation 0 = 0 holds true, which means we've proven the equality. Therefore, (a^2 - b^2)^2 = 4c^2(a^2 + b^2).