Question

Let \(S = \left\{\theta \in [-\pi, \pi] - \left\{\pm\frac{\pi}{2}\right\} : \sin\theta\tan\theta + \tan\theta = \sin 2\theta\right\}\). If \(T = \sum_{\theta \in S} \cos 2\theta\), then \(T + n(S)\) is equal to :

• A. \(7 + \sqrt{3}\)
• B. 9
• C. \(8 + \sqrt{3}\)
• D. 10

Hint:

Factor the equation as \(\sin\theta\,(2\sin\theta - 1)(\sin\theta + 1) = 0\) after multiplying by \(\cos\theta\); reject \(\sin\theta = -1\) (excluded values of theta).

Answer 1

The Correct Option is B

Since \(\theta \ne \pm\frac{\pi}{2}\), \(\cos\theta \ne 0\). Write the equation as \(\tan\theta(1 + \sin\theta) = 2\sin\theta\cos\theta\), i.e. \(\frac{\sin\theta(1 + \sin\theta)}{\cos\theta} = 2\sin\theta\cos\theta\).
Multiply by \(\cos\theta\): \(\sin\theta(1 + \sin\theta) = 2\sin\theta\cos^2\theta\).
\(\sin\theta\,[1 + \sin\theta - 2(1 - \sin^2\theta)] = 0 \Rightarrow \sin\theta\,(2\sin^2\theta + \sin\theta - 1) = 0 \Rightarrow \sin\theta\,(2\sin\theta - 1)(\sin\theta + 1) = 0.\)
Case \(\sin\theta = 0\): \(\theta = -\pi, 0, \pi\) (3 values).
Case \(\sin\theta = -1\): \(\theta = -\frac{\pi}{2}\), excluded.
Case \(\sin\theta = \frac12\): \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\) (2 values).
So n(S) = 5.
\(T = \cos(-2\pi) + \cos 0 + \cos 2\pi + \cos\frac{\pi}{3} + \cos\frac{5\pi}{3} = 1 + 1 + 1 + \frac12 + \frac12 = 4\).
\(T + n(S) = 4 + 5 = 9\).
Answer: (2) 9