Question

The value of

$\cos \left( 2\sin^{-1}\left(\frac{-1}{6}\right) \right) + \tan \left( \sec^{-1}\left(\frac{-13}{12}\right) \right)$ is

• A. $\frac{29}{36}$
• B. $\frac{19}{36}$
• C. $\frac{7}{36}$
• D. $\frac{49}{36}$

Answer 1

Answer: (B)

$\frac{19}{36}$

Answer 2

Let's evaluate the two terms separately.

Term 1: $\cos \left( 2\sin^{-1}\left(\frac{-1}{6}\right) \right)$

Let $\theta = \sin^{-1}\left(\frac{-1}{6}\right)$. Then $\sin(\theta) = -\frac{1}{6}$.

We need to find $\cos(2\theta)$. Using the double angle identity $\cos(2\theta) = 1 - 2\sin^2(\theta)$, we have:

$\cos(2\theta) = 1 - 2\left(-\frac{1}{6}\right)^2 = 1 - 2\left(\frac{1}{36}\right) = 1 - \frac{1}{18} = \frac{18 - 1}{18} = \frac{17}{18}$.

Term 2: $\tan \left( \sec^{-1}\left(\frac{-13}{12}\right) \right)$

Let $\phi = \sec^{-1}\left(\frac{-13}{12}\right)$. Then $\sec(\phi) = -\frac{13}{12}$.

We need to find $\tan(\phi)$. Using the identity $\tan^2(\phi) + 1 = \sec^2(\phi)$, we have:

$\tan^2(\phi) = \sec^2(\phi) - 1 = \left(-\frac{13}{12}\right)^2 - 1 = \frac{169}{144} - 1 = \frac{169 - 144}{144} = \frac{25}{144}$.

So, $\tan(\phi) = \pm \sqrt{\frac{25}{144}} = \pm \frac{5}{12}$. Since $\sec(\phi)$ is negative, $\phi$ is in the second quadrant, so $\tan(\phi)$ is negative.

Thus, $\tan(\phi) = -\frac{5}{12}$.

Now, we add the values of the two terms:

Value = $\frac{17}{18} + \left(-\frac{5}{12}\right) = \frac{17}{18} - \frac{5}{12} = \frac{34}{36} - \frac{15}{36} = \frac{19}{36}$.

Therefore, the value of the expression is $\frac{19}{36}$.