# Question 4a311

Aug 27, 2015

Trigonometric identities:
 tan (A+-B) = (tan A +- tan B)/(1 ""_+^(-) tan Atan B) #
${\sin}^{2} A + {\cos}^{A} = 1$
${\tan}^{2} A + 1 = {\sec}^{2} A = \frac{1}{\cos} ^ 2 A$
$\sin 2 A = \cos 2 \left(A - \frac{\pi}{2}\right) = - \cos 2 A$

Solution:
$\left(a + b\right) \tan \left(X - Y\right) = \left(a - b\right) \tan \left(X + Y\right)$
$\left(a + b\right) \left(\frac{\tan X - \tan Y}{1 + \tan X \tan Y}\right) = \left(a - b\right) \left(\frac{\tan X + \tan Y}{1 - \tan X \tan Y}\right)$
$\left(a + b\right) \left(\tan X - \tan Y\right) \left(1 - \tan X \tan Y\right) = \left(a - b\right) \left(\tan X + \tan Y\right) \left(1 + \tan X \tan Y\right)$
$a \tan Y \left({\tan}^{2} X + 1\right) = b \tan X \left({\tan}^{2} Y + 1\right)$
$a \sin 2 Y = b \sin 2 X$ ...(Eq. 1)

Now it is possible to determine that:
$a \cos 2 Y = b \cos 2 X$ ...(Eq. 2)

Summing the squares of equations 1 and 2:
${a}^{2} \left({\sin}^{2} 2 Y + {\cos}^{2} 2 Y\right) = {b}^{2} \left({\sin}^{2} 2 X + {\cos}^{2} 2 X\right)$
${a}^{2} = {b}^{2}$

From question:
$a \cos 2 Y + b \cos 2 X = c$
sum equation 1 to obtain:
$2 a \cos 2 Y = c$
Multiply by $c$ and add ${a}^{2} - {b}^{2} = 0$ to obtain:
${a}^{2} - {b}^{2} + {c}^{2} = 2 a c \cos 2 Y$