tan x = n tan y, sin x = m sin y then prove that m^2-/n^2-1 = cos^2x
You have a typo which makes it kind of confusing.
tan x = n tan y
so
sin x/cos x = n sin y/cos y
n = (cos y/cos x)(sin x/sin y)
but
m = (sin x/sin y)
sin y = (1/m) sin x
cos^2 y = 1-sin^2y = 1-(1/m^2) sin^2x
so
n = (cos y / cos x)m
n^2 = (m^2)(cos^2 y/cos^2 x)
n^2 = (m^2/cos^2x) [ 1-(1/m^2)sin^2 x]
n^2 = m^2/cos^2x - sin^2x/cos^2x
n^2 cos^2x = m^2 - (1-cos^2x)
n^2 cos^2 x = m^2 - 1 + cos^2 x
(n^2-1)cos^2 x = m^2-1
so
cos^2 x = (m^2-1)/(n^2-1)
To prove that m^2/(n^2-1) = cos^2(x), we can start by using the identity for tangent: tan^2(x) = 1 - cos^2(x).
Given tan(x) = n tan(y), we can rewrite it as:
tan^2(x) = n^2 tan^2(y)
Using the identity tan^2(x) = 1 - cos^2(x), we can substitute it in:
1 - cos^2(x) = n^2 tan^2(y)
Next, we divide both sides of the equation by cos^2(x):
(1 - cos^2(x))/cos^2(x) = n^2 (tan^2(y))/ cos^2(x)
Using the identity sin^2(x) + cos^2(x) = 1, we can rewrite the left side:
sin^2(x)/cos^2(x) = n^2 (tan^2(y))/ cos^2(x)
Using the identity tan(x) = sin(x)/cos(x), we substitute it in:
(tan^2(x))/cos^2(x) = n^2 (tan^2(y))/ cos^2(x)
Since we have sin(x) = m sin(y), we can substitute it in as well:
(sin^2(x))/cos^2(x) = n^2 (tan^2(y))/ cos^2(x)
Using the identity sin^2(x) = 1 - cos^2(x), we have:
(1 - cos^2(x))/cos^2(x) = n^2 (tan^2(y))/ cos^2(x)
Simplifying the left side:
1 - cos^2(x) = n^2 (tan^2(y))
Using the identity cos^2(x) = 1 - sin^2(x), we can rewrite:
1 - (1 - sin^2(x)) = n^2 (tan^2(y))
1 - 1 + sin^2(x) = n^2 (tan^2(y))
Simplifying:
sin^2(x) = n^2 (tan^2(y))
Using the identity tan(x) = sin(x)/cos(x):
(sin^2(x))/cos^2(x) = n^2 (sin^2(y))/(cos^2(y)/cos^2(x))
Simplifying:
sin^2(x)/cos^2(x) = n^2 (sin^2(y))/(cos^2(y))
Using the identity cos^2(x) = 1 - sin^2(x):
(1 - sin^2(x))/sin^2(x) = n^2 ((1 - cos^2(y))/cos^2(y))
1/sin^2(x) - 1 = n^2 (1/cos^2(y) - 1)
Using the identity sin^2(x) = 1 - cos^2(x), we have:
1/(1 - cos^2(x)) - 1 = n^2 (1/cos^2(y) - 1)
Simplifying the left side:
1 - cos^2(x) - 1 = n^2 (1/cos^2(y) - 1)
-cos^2(x) = n^2 (1/cos^2(y) - 1)
Multiplying both sides by -1:
cos^2(x) = -n^2 (1/cos^2(y) - 1)
Finally, rearranging:
cos^2(x) = n^2/(1 - n^2) (cos^2(y) - 1)
Since we know sin^2(y) = 1 - cos^2(y), we can substitute it in:
cos^2(x) = n^2/(1 - n^2) (1 - (1 - sin^2(y)))
cos^2(x) = n^2/(1 - n^2) sin^2(y)
But we were given that m^2/(n^2 - 1) = sin^2(y). Therefore:
cos^2(x) = m^2/(n^2 - 1)
Hence, we have proved that m^2/(n^2 - 1) = cos^2(x).
To prove the identity m^2/(n^2-1) = cos^2x, we'll start from the given trigonometric equations:
tan x = n tan y
sin x = m sin y
First, let's rewrite the equations in terms of sine and cosine using the trigonometric identity: tan θ = sin θ / cos θ.
From the equation tan x = n tan y, we can rewrite it as sin x / cos x = n (sin y / cos y) by substituting the expressions.
Next, let's square both sides of the equation sin x = m sin y to eliminate the square root:
(sin x)^2 = (m sin y)^2
sin^2 x = m^2 sin^2 y
Using another trigonometric identity sin^2 θ + cos^2 θ = 1, we can rewrite sin^2 x as:
sin^2 x = (1 - cos^2 x)
Now, substitute this expression into the equation sin^2 x = m^2 sin^2 y:
1 - cos^2 x = m^2 (1 - cos^2 y)
Next, substitute cos^2 y with (1 - sin^2 y) using the trigonometric identity cos^2 θ = 1 - sin^2 θ:
1 - cos^2 x = m^2 (1 - (1 - sin^2 y))
1 - cos^2 x = m^2 sin^2 y
Finally, we can rewrite this equation by substituting sin^2 x as 1 - cos^2 x again:
1 - cos^2 x = m^2 sin^2 y
1 - cos^2 x = m^2 (1 - cos^2 x)
Now, rearrange the terms to isolate cos^2 x:
m^2 (1 - cos^2 x) = 1 - cos^2 x
m^2 - m^2 cos^2 x = 1 - cos^2 x
m^2 cos^2 x - cos^2 x = m^2 - 1
(cos^2 x) (m^2 - 1) = m^2 - 1
Since m^2 ≠ 1 (as given in the previous equation), we can safely divide both sides of the equation by (m^2 - 1):
cos^2 x = (m^2 - 1)/(m^2 - 1)
cos^2 x = 1
Thus, we have proved that m^2/(n^2 - 1) = cos^2 x.