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Trigonometrical Equations
medium
यदि $3({\sec ^2}\theta + {\tan ^2}\theta ) = 5$, तो $\theta $ का व्यापक मान है
A
$2n\pi + \frac{\pi }{6}$
B
$2n\pi \pm \frac{\pi }{6}$
C
$n\pi \pm \frac{\pi }{6}$
D
$n\pi \pm \frac{\pi }{3}$
Solution
${\sec ^2}\theta + {\tan ^2}\theta = \frac{5}{3}$,
जबकि ${\sec ^2}\theta – {\tan ^2}\theta = 1$
$ \Rightarrow $ ${\tan ^2}\theta = \frac{1}{3} = {\tan ^2}\left( {\frac{\pi }{6}} \right) $
$\Rightarrow \theta = n\pi \pm \frac{\pi }{6}$.
Standard 11
Mathematics
