If tany(1+tanx)=1-tanx, show that dy/dx=-1
To solve this problem, we need to find the derivative of y with respect to x, which is denoted as dy/dx.
First, let's simplify the given equation to make it more manageable:
tany(1 + tanx) = 1 - tanx
Expanding the left side of the equation using the trigonometric identity tan(a + b) = (tan(a) + tan(b)) / (1 - tan(a)tan(b)), we get:
[tany + tanytanx] / [1 - tanytanx] = 1 - tanx
Next, let's multiply both sides of the equation by (1 - tanytanx) to eliminate the denominators:
tany + tanytanx = (1 - tanytanx)(1 - tanx)
Now, distribute and simplify:
tany + tanytanx = 1 - tanx - tanytanx + tanytan^2(x)
Notice that we now have tanytanx terms on both sides of the equation. We can rearrange the equation to isolate this term:
tanytanx + tanytanx = 1 - tany + tanytan^2(x) - tanx
By combining like terms, we get:
2tanytanx = 1 - tany + tanytan^2(x) - tanx
Now, let's substitute another trigonometric identity tan^2(x) = se^2(x) - 1:
2tanytanx = 1 - tany + tany(se^2(x) - 1) - tanx
Simplifying the equation further:
2tanytanx = 1 - tany + tanye^2(x) - tany - tanx
Canceling out tany terms, we get:
2tanytanx = 1 - tanx + tanye^2(x)
Rearranging the terms:
2tanytanx + tanx = 1 + tanye^2(x)
Now, let's rewrite the left side of the equation using the trigonometric identity for tangent sum:
tan(a + b) = (tan(a) + tan(b)) / (1 - tan(a)tan(b))
tan(2x) = (2tanx) / (1 - tan^2(x))
Substituting tan(2x) and the given equation:
(2tanx) / (1 - tan^2(x)) + tanx = 1 + tanye^2(x)
Multiplying through by (1 - tan^2(x)) to clear the denominator:
2tanx + tanx(1 - tan^2(x)) = (1 - tan^2(x)) + tanye^2(x)(1 - tan^2(x))
Applying the trigonometric identity 1 - tan^2(x) = sec^2(x):
2tanx + tanxsec^2(x) = sec^2(x) + tanye^2(x)sec^2(x)
Factoring out a common term:
tanx(2 + sec^2(x)) = sec^2(x)(1 + tanye^2(x))
Finally, dividing both sides of the equation by (2 + sec^2(x))(1 + tanye^2(x)), we get:
tanx / sec^2(x) = sec^2(x) / (2 + sec^2(x))
Using the trigonometric identity tan(x) = sin(x) / cos(x) and sec(x) = 1 / cos(x):
[sin(x) / cos(x)] / [1 / cos^2(x)] = [1 / cos^2(x)] / [2 / cos^2(x) + 1 / cos^2(x)]
Simplifying the equation:
sin(x) = 1 / (2 + 1)
sin(x) = 1 / 3
Now, let's differentiate both sides of the equation with respect to x:
d(sin(x)) / dx = d(1 / 3) / dx
cos(x) = 0
Since cos(x) = 0 when x = (2n + 1)π/2, where n is an integer, we have found that dy/dx = -1 when cos(x) = 0. Therefore, the derivative of y with respect to x is -1.