Question

If $\sin x + \cos x + \tan x + \cot x + \sec x + \csc x = 7$, then $\sin 2x = a - b \sqrt{7}; a, b \in \mathbb{N}$. The ordered pair $(a,b)$ can be equal to

• A. (22,8)
• B. (8,22)
• C. (22,2)
• D. (2,8)

Answer 1

Answer: (A)

(22,8)

Answer 2

Let $S = \sin x + \cos x$ and $P = \sin x \cos x$. The given equation simplifies to $S + \frac{1}{P} + \frac{S}{P} = 7$. This leads to $P = -\frac{S+1}{S-7}$. We also know $S^2 = 1 + 2P$, so $P = \frac{S^2-1}{2}$. Equating the expressions for $P$ gives $\frac{S^2-1}{2} = -\frac{S+1}{S-7}$. Since $S \neq -1$, we get $\frac{S-1}{2} = -\frac{1}{S-7}$, which simplifies to $S^2 - 8S + 9 = 0$. The solutions for $S$ are $4 \pm \sqrt{7}$. Since $S = \sin x + \cos x$ must be in the range $[-\sqrt{2}, \sqrt{2}]$, the only valid solution is $S = 4 - \sqrt{7}$. Then $\sin 2x = 2P = S^2 - 1 = (4 - \sqrt{7})^2 - 1 = (16 - 8\sqrt{7} + 7) - 1 = 23 - 8\sqrt{7} - 1 = 22 - 8\sqrt{7}$. Comparing with $\sin 2x = a - b\sqrt{7}$, we find $a=22$ and $b=8$. Thus, $(a,b) = (22,8)$.