I would like to know if there is a general method to solve equation looking like this:
$$\tan(\sec^{-1} 4)$$
without using a calculator (you have to find the exact value)?
How to proceed?
$\endgroup$ 42 Answers
$\begingroup$Imagine a right-angled triangle with one leg $k$ and hypotenuse $4k$ and angle $\theta$ between them. Then $\cos \theta = \frac{k}{4k}= \frac14$ and $\sec \theta = 4$, making $\sec^{-1}4 = \theta$.
The opposite leg is $\sqrt{(4k)^2-k^2}=\sqrt{15}k$ and so $\tan(\sec^{-1}4) = \tan \theta = \frac{\sqrt{15}k}{k}=\sqrt{15}$. Now you may need a calculator.
$\endgroup$ $\begingroup$Let $\sec^{-1}4=\theta\implies\sec\theta=4$
Now, $\tan^2\theta=\sec^2\theta-1$
Finally using the definition of the principal value of $\sec^{-1},0<\theta<\dfrac\pi2\implies\tan\theta>0$
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