1. ## Inverse trigonometry

${{tan}^{-1}} \frac{a}{x} + {{tan}^{-1}} \frac{b}{x} = \frac{x}{2}$

Find the value of x.
This doesn't seem to be striking me at all. I used the trigonometry identity of tan inverse A + tan inverse B, but the answer doesn't really spring from that..
Any help will be appreciated.

Also,
$tan({cos}^{-1}\frac{1}{5{\sqrt{2}}} - {sin}^{-1}\frac{4}{\sqrt{17}})$

Find the value of the expression here. I tried substituting and converting cos inverse to sin inverse but the identity is becoming too long and the answer ambiguous.

2. Originally Posted by SVXX
${{tan}^{-1}} \frac{a}{x} + {{tan}^{-1}} \frac{b}{x} = \frac{x}{2}$
Well, [tex]\theta= tan^{-1} \frac{b}{x}[\math] if and only if $\theta$ is an angle in a right triangle with "opposite side" of length a and "near side" of length x. Similarly, if $\phi= tan^{-1}\frac{b}{x}$ if and only if $\phi$ is an angle in a right triangle with "opposite side" b and "near side" x. Now, imagine those two triangles sharing the side of length x. You have a triangle with angle $\theta+ \phi= tan^{-1}\frac{a}{x}+ \frac{b}{x}= \frac{x}{2}$, base of lenght a+b and altitude of length x. Unfortunately, I don't really see where to go from there.

Find the value of x.
This doesn't seem to be striking me at all. I used the trigonometry identity of tan inverse A + tan inverse B, but the answer doesn't really spring from that..
Any help will be appreciated.

Also,
$tan({cos}^{-1}\frac{1}{5{\sqrt{2}}} - {sin}^{-1}\frac{4}{\sqrt{17}})$

Find the value of the expression here. I tried substituting and converting cos inverse to sin inverse but the identity is becoming too long and the answer ambiguous.[/QUOTE]

3. Hello, SVXX!

Here's the second one . . .

Simplify: . $\tan\left[\cos^{-1}\!\left(\tfrac{1}{5{\sqrt{2}}}\right) - \sin^{-1}\!\left(\tfrac{4}{\sqrt{17}}\right)\right]$
$\text{We have: }\;\tan\left[\underbrace{\cos^{-1}\!\left(\tfrac{1}{5\sqrt{2}}\right)}_{\alpha} - \underbrace{\sin^{-1}\!\left(\tfrac{4}{\sqrt{17}}\right)}_{\beta}\rig ht]$ .[1]

$\alpha \:=\:\cos^{-1}\left(\tfrac{1}{5\sqrt{2}}\right) \quad\Rightarrow\quad \cos\alpha \:=\:\tfrac{1}{5\sqrt{2}} \:=\:\frac{adj}{hyp} \quad\Rightarrow\quad opp \:=\:7$
. . $\tan\alpha \:=\:\frac{opp}{adj} \:=\:\tfrac{7}{1} \:=\:7 \quad\Rightarrow\quad \boxed{\alpha \:=\:\tan^{-1}(7)}$

$\beta \:=\:\sin^{-1}\left(\frac{4}{\sqrt{17}}\right) \quad\Rightarrow\quad \sin\beta \:=\:\frac{4}{\sqrt{17}} \:=\:\frac{opp}{hyp} \quad\Rightarrow\quad adj \:=\:1$
. . $\tan\beta \:=\:\frac{opp}{adj} \:=\:\frac{4}{1} \:=\:4 \quad\Rightarrow\quad \boxed{\beta \:=\:\tan^{-1}(4)}$

Substitute into [1]:

$\tan[\alpha - \beta] \;=\;\tan\left[\tan^{-1}(7) - \tan^{-1}(4)\right]$

. . . . . . . $= \;\frac{\tan[\tan^{-1}(7)] - \tan[\tan^{-1}(4)]}{1 - [\tan^{-1}(7)][\tan^{-1}(4)]}$

. . . . . . . $= \;\frac{7-4}{1-7\!\cdot\!4} \;=\;\frac{\text{-}3}{\text{-}27} \;=\;\boxed{\frac{1}{9}}$

4. Ah mighty thanks Soroban! So the trick was to directly convert both into inverse tan.