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Question: Determine exact value of $\tan ({\sec }^{-1}(-\sqrt{26}))$

1. Set $\theta = \sec^{-1}(-\sqrt{26})

2. Find what quadrant $\theta$ is in

C

A

S

T

First, find out what ${\sec }^{-1}$ is

$\sec (x)=\frac{1}{\cos (x)}$

$x=\sec (y)$

$x=\frac{1}{\cos (y)}$

$\frac{1}{x}=\cos (y)$

$\arccos (\frac{1}{x})=y$

${\sec }^{-1}(x)=\arccos (\frac{1}{x})$

$\sec (\theta )=-\sqrt{26}$

Remember that $\theta = \sec^{-1}(-\sqrt{26})

$\frac{1}{\cos (\theta )}=-\sqrt{26}$

$\cos (\theta )=\frac{1}{-\sqrt{26}}$

C

A

S

T

Theta is in these two quadrants

Remember, that ${\sec }^{-1}$ is only defined on $[0,\frac{\pi }{2})U(\frac{\pi }{2},\pi ]$

C

A

S

T

So, theta can only occupy $(\frac{\pi }{2},\pi ]$

C

A

S

T

3. Take cos of both sides

$\theta = \sec^{-1}(-\sqrt{26})

$\cos (\theta )=\frac{1}{-\sqrt{26}}$

4. Draw a triangle

$\theta$

$-1$

$\sqrt{26}$

$5$

5. Determine $\tan (\theta )$

$\tan (\theta )=\frac{opp}{adj}$

$\tan (\theta )=\frac{5}{1}$

$\tan (\theta )=5$