Solving for $t$ in $A \sec^2(t) + (1-A)\sec^2(at) = B$ 
Let $0 < A < 1$, $a >0$, $B > 1$ and suppose that the following identity holds
$$A \sec^2(t) + (1-A)\sec^2(at) = B$$
Express $t$ in terms of $a, A,$ and $B$

This doesn't arise in any specific context it's just fallen out of a long computation I've been doing trying to get a critical point of something. I've tried using all the trig identities I know and I know that there exists a $t$ that solves this but I can't find it explicitly.
(I've edited it to say that $B > 1$ otherwise there's no solution)
 A: Without any restrictions, I do not think that this (interesting) equation could be solved analytically and numerical methods would be required (knowing that there is an infinite number of symmetric solutions).
In any manner, I should consider instead the problem of the zero's of function
$$f_a(x)=A \cos ^2(a t)-B \cos ^2(t) \cos ^2(a t)-A \cos ^2(t)+\cos ^2(t)$$ in order to avoid the vertical asymptotes.
If $a$ is an integer, this reduces to a polynomial in $x=\cos^{-1}(t)$ but the degree is high. For example,
$$f_2(x)=-A+ (5 A+B-1)x^2-4  (A+B)x^4+4 B x^6$$ is a cubic equation in $x^2$ and can be solved. The same for
$$f_3(x)=-(8 A+1) x^2+ (24 A+9 B)x^4- 8(2 A+3 B)x^6+16 B x^8$$ In any of this cases, I wish you a lot of fun with radicals.
If we are concernd by the smallest root, we could (may be !) approximate the solution using Taylor series built at $t=0$. This would give
$$(1-B)+\alpha\, t^2 +\beta\, t^4+\gamma t^6+O\left(t^8\right)$$ where
$$\alpha=A+B+a^2(B-A)-1$$
$$\beta=\frac{1}{3} \left(\left(a^4-1\right) A-\left(a^4+3 a^2+1\right) B+1\right)$$
$$\gamma=\frac{1}{45} \left(2 \left(1-a^6\right) A+\left(a^2+1\right) \left(2 a^4+13
   a^2+2\right) B-2\right)$$
Just to give a try with $A=0.25$, $B=9.87$, $a=2.34$. Solving the cubic would give $t=0.503545$ while the solution is $0.549738$. Not fantastic at all !
So, use Newton method starting with $t_0=0$.
