How do you evaluate $sec (2 arcsin (- \frac{1}{3}))$ $sec(2 arcsin(-1/3))$ ?

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Answer 1 · 2016-07-19T03:25:21

$\frac{9}{7}$ $9/7$

Let $a = a r c sin (- \frac{1}{3})$ $a = arc sin (-1/3)$ . Then,

$sin a = (- \frac{1}{3})$ $sin a = (-1/3)$ .

So, $cos 2 a = 1 - 2 {sin}^{2} a = 1 - 2 (\frac{1}{9}) = \frac{7}{9}$ $cos 2 a=1-2 sin^2 a= 1 - 2 (1/9)=7/9$

The given expression is $sec 2 a = \frac{1}{cos 2 a} = \frac{9}{7}$ $sec 2 a=1/(cos 2a) =9/7$ ..

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