I try to find the values of W that met the equation
$(1-n)*\arctan(a*W)-n*\arctan(b*W)=\theta $
in which $0<n<1$, a and b are positive reals, and $\theta$ is an angle such that $-\pi/2<\theta<\pi/2$. would anyone know a trigonometric formula or a solution for such a problem? Thank you.
I have the feeling that, for the most general case, an explicit solution does not exist and then you should use a numerical method for finding the zero of function $$f(W)=(1-n) \tan ^{-1}(a W)+n \tan ^{-1}(b W)-\theta$$ Its derivative $$f'(W)=\frac{a (1-n)}{1+a^2 W^2}+\frac{b n}{1+b^2 W^2}$$ is always positive because of the conditions $(a >0, b>0, 0 < n <1)$ so $f(W)$ is an increasing function and there is only one root.
You must notice that $f(0)=-\theta$. So, if $\theta >0$ the solution will be positive and if $\theta <0$ the solution will be negative.
If you use Newton method, the first iterate, starting with $W_0=0$, will be $$W_1=\frac{\theta }{a (1-n)+b n}$$ which will be updated according to $$W_{n+1}=W_n-\frac{f(W_n)}{f'(W_n)}$$
For illustration, let us use $n=0.2345$, $a=1$, $b=\pi$ and $\theta=\frac \pi 3$ with $W_0=0$. Newton iterates would be $$\left( \begin{array}{cc} n & W_n \\ 0 & 0 \\ 1 & 0.6971077 \\ 2 & 1.1848712 \\ 3 & 1.3874361 \\ 4 & 1.4120731 \\ 5 & 1.4123810 \\ 6 & 1.4123811 \end{array} \right)$$