Trigonometry puzzle

From the first equation we have that

$\cos^{2}(A) = \tan^{2}(B) \Longrightarrow 1 - \sin^{2}(A) = \sec^{2}(B) - 1$

$\Longrightarrow \sin^{2}(A) = 2 - \sec^{2}(B)$ .

From the second equation we have that

$\sec^{2}(B) = \cot^{2}(C) = \dfrac{\cos^{2}(C)}{1 - \cos^{2}(C)}$ .

Combining these two results with the third equation, we have that

$\sin^{2}(A) = 2 - \dfrac{\tan^{2}(A)}{1 - \tan^{2}(A)} = 2 - \dfrac{\sin^{2}(A)}{1 - 2\sin^{2}(A)}$

$\Longrightarrow \sin^{2}(A) * (1 - 2\sin^{2}(A)) = 2 - 5\sin^{2}(A)$

$\Longrightarrow 2\sin^{4}(A) - 6\sin^{2}(A) + 2 = 0 \Longrightarrow \sin^{4}(A) - 3\sin^{2}(A) + 1 = 0$ ,

which is quadratic in $\sin^{2}(A)$ . So using the quadratic formula we have that

$\sin^{2}(A) = \dfrac{3 \pm \sqrt{5}}{2}$ .

Now $0 \le \sin^{2}(A) \le 1$ , so we are left with $\sin^{2}(A) = \dfrac{3 - \sqrt{5}}{2}$ .

Now suppose that $\sin(A) = a + b\sqrt{5}$ . Then

$\sin^{2}(A) = a^{2} + 5b^{2} + 2ab\sqrt{5}$ .

Comparing this to the value we found previously, we see that

(i) $a^{2} + 5b^{2} = \dfrac{3}{2}$ and (ii) $2ab = -\dfrac{1}{2}$ .

Substituting (ii) into (i) leads to the result

$16a^{4} - 24a^{2} + 5 = 0 \Longrightarrow a^{2} = \dfrac{1}{4}$ or $a^{2} = \dfrac{5}{4}$ .

From the first of these solutions we have that $a = \pm \dfrac{1}{2}$ . From equation (ii) this implies that $b = \pm \dfrac{1}{2}$ . Testing out the $4$ options, we see that

$a = -\dfrac{1}{2}, b = \dfrac{1}{2}$ , and so $\sin(A) = \boxed{\dfrac{\sqrt{5} - 1}{2}}$ .

Note: The solution $a^{2} = \dfrac{5}{4}$ gives us $a = \pm \dfrac{\sqrt{5}}{2}, b = \pm \dfrac{\sqrt{5}}{10}$ , from which $a = \dfrac{\sqrt{5}}{2}, b = -\dfrac{\sqrt{5}}{10}$ yields the same final answer as found above.

Comment: Note that we could cheat a bit and just let $A = B = C$ to find that

$\cos(A) = \tan(A) \Longrightarrow \cos^{2}(A) = \sin(A) \Longrightarrow \sin^{2}(A) + \sin(A) - 1 = 0$ ,

from which we find that $\sin(A) = \dfrac{-1 \pm \sqrt{5}}{2}$ .

Now since $-1 \le \sin(A) \le 1$ we conclude that $\dfrac{\sqrt{5} - 1}{2}$ is the answer in this special case. What I showed above, however, was that this is the value of $\sin(A)$ in general for any $A, B, C$ that satisfy the three initial equations.