<s>[INST] Evaluating the expression $\cos(\frac{1}{2}\tan^{-1}(-\frac{4}{3}))$ I can prove that $\cos(\tan^{-1}(x)) = +\dfrac{1}{\sqrt{1+x^2}}$, set $y=\tan^{-1}(x)$ and therefore $y=\cos^{-1}(\frac{1}{\sqrt{1+x^2}})$
Dividing both sides by $2$: 
$\dfrac{y}{2}=\dfrac{1}{2}\cos^{-1}(\frac{1}{\sqrt{1+x^2}})$ 
$\cos(\dfrac{1}{2}\tan^{-1}(x))=\cos(\dfrac{1}{2}\cos^{-1}(\frac{1}{\sqrt{1+x^2}}))$ 
$\cos(\dfrac{1}{2}\tan^{-1}(-\frac{4}{3}))=\cos(\dfrac{1}{2}\cos^{-1}(\frac{3}{5}))$ 
This is my approach to the problem, are there any better ways to simplify the initial expression and perhaps find an approximation even without using a calculator?
 [/INST] Consider the following :
$$\begin{align}
&\implies\frac{\arctan\left(\frac{-4}{3}\right)}{2}=y \\
&\implies2y=\arctan(\frac{-4}{3})\\
&\implies\tan(2y)=-4/3\\
&\implies\frac{2\tan y}{1-\tan^2y}=-4/3\\ 
&\implies2\tan^2y-3\tan y-2=0\\
\end{align}$$
Solving this equation we get-
$$\tan y=-1/2$$and $$\tan y=2$$
Using this we can find the value of $\cos y$ which will give the value of the expression. It comes out to be:
$$\cos y=\frac{1}{\sqrt{5}}$$
Hope my answer is helpful. Sorry for any errors and please let me know if there are some. 
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