Question

$\tan(\cos^{-1}(\frac{3}{\sqrt{10}})) + 2\sin^{-1}(\frac{4}{\sqrt{17}})$

Answer 1

$\frac{1}{3} + \pi - \tan^{-1}(\frac{8}{15})$

Answer 2

To evaluate the expression $\tan(\cos^{-1}(\frac{3}{\sqrt{10}})) + 2\sin^{-1}(\frac{4}{\sqrt{17}})$, we will evaluate each term separately.

Part 1: $\tan(\cos^{-1}(\frac{3}{\sqrt{10}}))$

Let $\theta = \cos^{-1}(\frac{3}{\sqrt{10}})$. This implies $\cos \theta = \frac{3}{\sqrt{10}}$. Since $\frac{3}{\sqrt{10}} > 0$, $\theta$ lies in the first quadrant, i.e., $0 < \theta < \frac{\pi}{2}$.

We can construct a right-angled triangle where the adjacent side is 3 and the hypotenuse is $\sqrt{10}$. Using the Pythagorean theorem, the opposite side $y$ is: $y^2 + 3^2 = (\sqrt{10})^2$ $y^2 + 9 = 10$ $y^2 = 1$ $y = 1$ (since length must be positive).

Now, we can find $\tan \theta$: $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{1}{3}$. So, $\tan(\cos^{-1}(\frac{3}{\sqrt{10}})) = \frac{1}{3}$.

Part 2: $2\sin^{-1}(\frac{4}{\sqrt{17}})$

Let $\phi = \sin^{-1}(\frac{4}{\sqrt{17}})$. This implies $\sin \phi = \frac{4}{\sqrt{17}}$. Since $\frac{4}{\sqrt{17}} > 0$, $\phi$ lies in the first quadrant, i.e., $0 < \phi < \frac{\pi}{2}$.

We can construct a right-angled triangle where the opposite side is 4 and the hypotenuse is $\sqrt{17}$. Using the Pythagorean theorem, the adjacent side $x$ is: $x^2 + 4^2 = (\sqrt{17})^2$ $x^2 + 16 = 17$ $x^2 = 1$ $x = 1$ (since length must be positive).

Now, we can find $\tan \phi$: $\tan \phi = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{4}{1} = 4$. So, $\phi = \tan^{-1}(4)$.

We need to evaluate $2\phi = 2\tan^{-1}(4)$. We use the formula for $2\tan^{-1}x$: $2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ if $-1 < x < 1$. $2\tan^{-1}x = \pi + \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ if $x > 1$. $2\tan^{-1}x = -\pi + \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ if $x < -1$.

In our case, $x=4$, which is greater than 1. So we use the second formula: $2\tan^{-1}(4) = \pi + \tan^{-1}\left(\frac{2 \times 4}{1 - 4^2}\right)$ $= \pi + \tan^{-1}\left(\frac{8}{1 - 16}\right)$ $= \pi + \tan^{-1}\left(\frac{8}{-15}\right)$ Since $\tan^{-1}(-y) = -\tan^{-1}(y)$: $2\tan^{-1}(4) = \pi - \tan^{-1}\left(\frac{8}{15}\right)$.

Combining the parts:

The original expression is: $\tan(\cos^{-1}(\frac{3}{\sqrt{10}})) + 2\sin^{-1}(\frac{4}{\sqrt{17}})$ $= \frac{1}{3} + \left(\pi - \tan^{-1}\left(\frac{8}{15}\right)\right)$ $= \frac{1}{3} + \pi - \tan^{-1}\left(\frac{8}{15}\right)$

This is the simplified form of the expression.