How to evaluate inverse trigonometric functions in exponents? I recently started working on inverse trigonometry, I have done many problems that include conversion of inverse functions and many more formulas but how should we approach a question when the inverse functions are given in exponents.
For example, I came to a question,


  $12^{\arcsin(x)} + 12^{\arccos(x)} + 12^{\arctan(x)} &gt;3\cdot k^{\pi/k}$
  
  Find $k$?


Can you please provide me a start. I will solve the rest on my own.
Thanks.
I know no general method to approach a question when the inverse functions are given in exponents, but I guess that in general such questions hide simple ideas under a cumbersome form with artificially chosen functions and coefficients.. For instance, in this question the inequality holds for all positive $k$. We can show this as follows. 

Since $\operatorname{arcsin} x+\operatorname{arccos} x=\pi/2$ for each $x$, 

$$12^{\operatorname{arcsin} x}+12^{\operatorname{arccos} x}+12^{\operatorname{arctan} x}&gt;$$ 
$$12^{\operatorname{arcsin} x}+12^{\operatorname{arccos} x}\ge \mbox{ (by AM-GM) }$$ $$
12^{(\operatorname{arcsin} x+\operatorname{arccos} x)/2}=2\cdot 12^{\pi/4}\approx 14.080.$$

On the other hand, the derivative of the function $3k^{\pi/k}$ is $\frac{3\pi k^{\pi/k}}{k^2}(1-\ln k)$. Therefore the function increases when $0&lt;k&lt;e$, attains its maximum $3e^{\pi/e}\approx 9.529$ at $k=e$, and decreases when $k&gt;e$. 

PS. The following graph (drawn by Mathcad) suggests that the minimum value of a function
$f(x)=12^{\operatorname{arcsin} x}+12^{\operatorname{arccos} x}+12^{\operatorname{arctan} x}$ 
is about $18.5237$ and is attained when $x$ is a solution of a transcendental equation $f’(x)=0$, that is when

$$(12^{\operatorname{arccos} x}-12^{\operatorname{arcsin} x})(1+x^2)=12^{\operatorname{arctan} x}\sqrt{1-x^2}.$$
You can try also to convert arcsin and arccos to their complex log expressions. If do that, the x goes down, out of the exponents.