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Trigonometrical Equations
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If $\frac{{1 - {{\tan }^2}\theta }}{{{{\sec }^2}\theta }} = \frac{1}{2}$, then the general value of $\theta $ is
A
$n\pi \pm \frac{\pi }{6}$
B
$n\pi + \frac{\pi }{6}$
C
$2n\pi \pm \frac{\pi }{6}$
D
None of these
Solution
(a) $\frac{{1 – {{\tan }^2}\theta }}{{{{\sec }^2}\theta }} = \frac{1}{2} $
$\Rightarrow {\cos ^2}\theta – {\sin ^2}\theta = \frac{1}{2}$
$ \Rightarrow $ $\cos 2\theta = \frac{1}{2} = \cos \left( {\frac{\pi }{3}} \right)$
$ \Rightarrow $ $2\theta = 2n\pi \pm \frac{\pi }{3} $
$\Rightarrow \theta = n\pi \pm \frac{\pi }{6}$.
Standard 11
Mathematics
