# Question #7a9c0

Jul 22, 2017

${\sin}^{-} 1 x + {\tan}^{-} 1 x = \frac{\pi}{2}$

$\implies {\tan}^{-} 1 x = \frac{\pi}{2} - {\sin}^{-} 1 x$

$\implies {\tan}^{-} 1 x = {\cos}^{-} 1 x$

$\implies {\tan}^{-} 1 x = {\tan}^{-} 1 \left(\frac{\sqrt{1 - {x}^{2}}}{x}\right)$

$\implies x = \left(\frac{\sqrt{1 - {x}^{2}}}{x}\right)$

$\implies {x}^{2} = \frac{1 - {x}^{2}}{x} ^ 2$

$\implies {x}^{4} = 1 - {x}^{2}$

$\implies {x}^{4} + {x}^{2} - 1 = 0$

$\implies {x}^{2} = \frac{- 1 + \sqrt{{\left(- 1\right)}^{2} - 4 \cdot 1 \left(- 1\right)}}{2}$

$\implies {x}^{2} = \frac{\sqrt{5} - 1}{2}$

$\implies x = \sqrt{\frac{\sqrt{5} - 1}{2}}$

Jul 22, 2017

#### Explanation:

Let $u = S {\in}^{- 1} x$.
Let $v = T a {n}^{- 1} x$.

Since $\sin u = x$, we may draw a right triangle having $x$ as the length of the side opposite angle u, and $1$ as the length of the hypotenuse. This we do because $\sin = \frac{o p p}{h y p}$. By the Pythagorean Theorem, the length of the remaining leg of that triangle is $\sqrt{1 - {x}^{2}}$.

Since $\tan v = x$, we may draw a right triangle having $x$ as the length of the side opposite angle v, and $1$ as the length of the side adjacent to angle v. This we do because $\tan = \frac{o p p}{a \mathrm{dj}}$. By the Pythagorean Theorem, the length of the remaining leg of that triangle is $\sqrt{1 + {x}^{2}}$.

Now examine $S {\in}^{- 1} x + T a {n}^{- 1} x = \frac{\pi}{2}$.
We take the cosine of both sides, since $\cos \left(\frac{\pi}{2}\right) = 0$.
$\cos \left(S {\in}^{- 1} x + T a {n}^{- 1} x\right) = \cos \left(\frac{\pi}{2}\right)$
$\cos \left(S {\in}^{- 1} x + T a {n}^{- 1} x\right) = 0$
Using u and v, we have:
$\cos \left(u + v\right) = 0$
$\cos u \cos v - \sin u \sin v = 0$

From the first triangle that we drew, above, we see that
$\cos u = \sqrt{1 - {x}^{2}}$ and $\sin u = x$.
From the second triangle we see that
$\cos v = \frac{1}{\sqrt{1 + {x}^{2}}}$ and $\sin v = \frac{x}{\sqrt{1 + {x}^{2}}}$.

$\cos u \cos v - \sin u \sin v = 0$
$\left(\sqrt{1 - {x}^{2}}\right) \left(\frac{1}{\sqrt{1 + {x}^{2}}}\right) - \left(x\right) \left(\frac{x}{\sqrt{1 + {x}^{2}}}\right) = 0$

$\frac{\sqrt{1 - {x}^{2}}}{\sqrt{1 + {x}^{2}}} - {x}^{2} / \sqrt{1 + {x}^{2}} = 0$
Multiply by the denominator, which is nonzero:
$\sqrt{1 - {x}^{2}} - {x}^{2} = 0$

Now for some algebra:
$\sqrt{1 - {x}^{2}} = {x}^{2}$
$1 - {x}^{2} = {x}^{4}$
$0 = {x}^{4} + {x}^{2} - 1$
${x}^{2} = \frac{- 1 \pm \sqrt{5}}{2}$
Since ${x}^{2}$ must be positive, replace $\pm$ with +
${x}^{2} = \frac{- 1 + \sqrt{5}}{2}$
Now, the values of both $\arcsin x$ and $\arctan x$ are negative when $x < 0$. Therefore, the negative root is not a solution to the equation. The only solution is:
$x = \sqrt{\frac{- 1 + \sqrt{5}}{2}}$