Question

Show that eliminating x & y from the equations, sin x + sin y = a; cos x + cos y = b & tanx + tany = c gives $\frac{8ab}{(a^{2}+b^{2})^{2}-4a^{2}}=c$.

Answer 1

$\frac{8ab}{(a^{2}+b^{2})^{2}-4a^{2}}=c$

Answer 2

Given equations:

1. $\sin x + \sin y = a$
2. $\cos x + \cos y = b$
3. $\tan x + \tan y = c$

From (1) and (2), square and add them: $(\sin x + \sin y)^2 + (\cos x + \cos y)^2 = a^2 + b^2$ $(\sin^2 x + \sin^2 y + 2 \sin x \sin y) + (\cos^2 x + \cos^2 y + 2 \cos x \cos y) = a^2 + b^2$ $(\sin^2 x + \cos^2 x) + (\sin^2 y + \cos^2 y) + 2(\sin x \sin y + \cos x \cos y) = a^2 + b^2$ $1 + 1 + 2 \cos(x-y) = a^2 + b^2$ $2 + 2 \cos(x-y) = a^2 + b^2$ $\cos(x-y) = \frac{a^2+b^2-2}{2}$ (Eq. 4)

From (1) and (2), use sum-to-product formulas: $2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) = a$ $2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) = b$

Divide the first by the second: $\tan\left(\frac{x+y}{2}\right) = \frac{a}{b}$ (Eq. 5)

Now, let's work with equation (3): $\tan x + \tan y = c$ $\frac{\sin x}{\cos x} + \frac{\sin y}{\cos y} = c$ $\frac{\sin x \cos y + \cos x \sin y}{\cos x \cos y} = c$ $\frac{\sin(x+y)}{\cos x \cos y} = c$ (Eq. 6)

We need to express $\sin(x+y)$ and $\cos x \cos y$ in terms of $a$ and $b$.

From (5), we can find $\sin(x+y)$ and $\cos(x+y)$ using half-angle identities: $\sin(x+y) = \frac{2 \tan\left(\frac{x+y}{2}\right)}{1 + \tan^2\left(\frac{x+y}{2}\right)} = \frac{2(a/b)}{1 + (a/b)^2} = \frac{2a/b}{(b^2+a^2)/b^2} = \frac{2ab}{a^2+b^2}$ (Eq. 7) $\cos(x+y) = \frac{1 - \tan^2\left(\frac{x+y}{2}\right)}{1 + \tan^2\left(\frac{x+y}{2}\right)} = \frac{1 - (a/b)^2}{1 + (a/b)^2} = \frac{(b^2-a^2)/b^2}{(b^2+a^2)/b^2} = \frac{b^2-a^2}{a^2+b^2}$ (Eq. 8)

Now, let's find $\cos x \cos y$. We use the identity $2 \cos x \cos y = \cos(x+y) + \cos(x-y)$. Substitute (4) and (8) into this identity: $2 \cos x \cos y = \frac{b^2-a^2}{a^2+b^2} + \frac{a^2+b^2-2}{2}$ $2 \cos x \cos y = \frac{2(b^2-a^2) + (a^2+b^2)(a^2+b^2-2)}{2(a^2+b^2)}$ $2 \cos x \cos y = \frac{2b^2-2a^2 + (a^2+b^2)^2 - 2(a^2+b^2)}{2(a^2+b^2)}$ $2 \cos x \cos y = \frac{(a^2+b^2)^2 + 2b^2-2a^2 - 2a^2-2b^2}{2(a^2+b^2)}$ $2 \cos x \cos y = \frac{(a^2+b^2)^2 - 4a^2}{2(a^2+b^2)}$ So, $\cos x \cos y = \frac{(a^2+b^2)^2 - 4a^2}{4(a^2+b^2)}$ (Eq. 9)

Finally, substitute (7) and (9) into (6): $c = \frac{\sin(x+y)}{\cos x \cos y} = \frac{\frac{2ab}{a^2+b^2}}{\frac{(a^2+b^2)^2 - 4a^2}{4(a^2+b^2)}}$ $c = \frac{2ab}{a^2+b^2} \times \frac{4(a^2+b^2)}{(a^2+b^2)^2 - 4a^2}$ $c = \frac{8ab}{(a^2+b^2)^2 - 4a^2}$

This proves the required relation.