How to evaluate $\tan^{-1}\left(\frac4x\right)-\tan^{-1}\left(\frac3x\right)$ This is what I did but please correct me if I am wrong.
Let $y=\tan^{-1}\Big(\dfrac{4}{x}\Big)-\tan^{-1}\Big(\dfrac{3}{x}\Big)$.
$\tan(y)\\
=\dfrac{\tan\Big(\tan^{-1}\Big(\dfrac{4}{x}\Big)\Big)-\tan\Big(\tan^{-1}\Big(\dfrac{3}{x}\Big)\Big)}{1-\tan\Big(\tan^{-1}\Big(\dfrac{4}{x}\Big)\Big)\tan\Big(\tan^{-1}\Big(\dfrac{3}{x}\Big)\Big)}\\
=\dfrac{\dfrac{4}{x}-\dfrac{3}{x}}{1-\dfrac{4}{x}\times\dfrac{3}{x}}\\
=\dfrac{\dfrac{1}{x}}{1-\dfrac{12}{x^2}}\\
=\dfrac{x}{x^2-12}
$
 A: Yes, your approach is correct but it must be completed. First of all, what you found is $\tan(y)$ in the end so with the correct formula $$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a\tan b}$$
the answer becomes $$y = \tan^{-1}\bigg(\frac{x}{x^2+12}\bigg)$$
And here is another approach by using geometry: 

Here, notice that $\tan(a) = \frac{3}{x}$ and $\tan(b) = \frac{4}{x}$ and what we are seeking is $\angle ACD = c$. Then, if we use Law of Sines for the triangle $ADC$, we have an equality:
$$\frac{|AD|}{\sin(\angle DCA)} = \frac{|DC|}{\sin(\angle DAC)}$$
which is
$$\frac{1}{\sin(c)} = \frac{\sqrt{x^2+9}}{\frac{x}{\sqrt{x^2+16}}}$$
therefore we have $\sin(c) = \frac{x}{\sqrt{x^4+25x^2+144}}$. Now we can construct another right angle triangle with one of its angle is $c$, also satisfying this condition:

Then notice that by Pythagorean Theorem, 
$$|LM|^2 = x^4+25x^2+144-x^2 = x^4+24x^2+144 = (x^2+12x)^2 \implies |LM| = x^2+12x$$
Finally, $$\tan(c) = \frac{|KL|}{|LM|} = \frac{x}{x^2+12} \implies c = \tan^{-1}\bigg(\frac{x}{x^2+12}\bigg)$$
A: Use the formula $$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a\tan b}\ $$ with $tan(a)=4/x$ and $tan(b)=3/x$ to get $$tan(a-b)=x/(x^2+12)$$ Therefore the answer is $$tan^{-1}(x/(x^2+12))$$
