if tanA=ntanX
sinA=msinX
then prove that cos^2A=(m^2-1)/(n^2-1)
To prove that cos^2A = (m^2-1)/(n^2-1), we will start by rewriting the given trigonometric equations in terms of sine and cosine.
From the given information:
1) tanA = n tanX
2) sinA = m sinX
We know that tanA = sinA / cosA and tanX = sinX / cosX. So we can rewrite equation 1 as:
sinA / cosA = n (sinX / cosX)
Cross-multiplying gives us:
sinA * cosX = n * sinX * cosA
Similarly, we can rewrite equation 2 as:
sinA = m sinX
Dividing equation 2 by equation 1, we have:
(m sinX) / (sinA * cosX) = (n sinX * cosA) / (sinA * cosX)
Canceling out sinX and cosX terms gives us:
m / sinA = n * cosA
Rearranging the equation, we get:
(m / n) = cosA / sinA
Now, using the identity cos^2A + sin^2A = 1, we can replace cosA / sinA with its equivalent terms:
(m / n) = cosA / sinA
(m / n) = cosA / (sqrt(1 - cos^2A))
(m^2 / n^2) = (cosA)^2 / (1 - cos^2A)
Cross-multiplying, we obtain:
(m^2 - m^2 * cos^2A) = n^2 * (cosA)^2
Rearranging the terms:
m^2 - m^2 * cos^2A = n^2 * cos^2A
m^2 = m^2 * cos^2A + n^2 * cos^2A
m^2 = cos^2A * (m^2 + n^2)
Now, isolating cos^2A, we divide both sides by (m^2 + n^2):
m^2 / (m^2 + n^2) = cos^2A
1 - n^2 / (m^2 + n^2) = cos^2A
Simplifying the right side, we get:
(m^2 - n^2) / (m^2 + n^2) = cos^2A
Finally, replacing m^2 - n^2 with (m^2 - 1) and m^2 + n^2 with (n^2 - 1), we obtain:
cos^2A = (m^2 - 1) / (n^2 - 1)
Therefore, we have proved that cos^2A = (m^2 - 1) / (n^2 - 1).