Generator of a rotation matrix $$T(\phi)=
\begin{bmatrix}
    \cos(\theta)                &\sin(\theta)              & 0  \\
    \sin(\theta)\cos(\phi)       & -\cos(\theta)\cos(\phi)   & \sin(\phi)\\
    \sin(\theta)\sin(\phi)       & -\cos(\theta)\sin(\phi)   &-\cos(\phi)
\end{bmatrix}$$
Here the matrix $T$ is parametrized by $\phi$ and $\theta$=some constant angle. Can I find out the generators of this orthogonal transformation parametrized by the angle $\phi$ ? 
If my approach is wrong, how do I find the generators of this matrix and exponentiate it?
I have  derived an infinitesimal transformation which leads to $$T(\delta\phi)=\bigg[I+t\delta\phi\bigg]$$
where $t$ is,
$$t=
\begin{bmatrix}
    0                &0              & 0  \\
    0       & 0   & 1\\
    \sin(\theta)       & -\cos(\theta)   &0
\end{bmatrix}.$$
Here $\theta$ is some fixed angle, say $120$ degrees or $69$ degrees  or anything, but it remains constant. Can I exponentiate this matrix to get $$e^{-\hat{t}\phi}$$
Is it correct? Where am I going wrong if I am completely incorrect?
Edit: if $\theta$ is a fixed constant, there is no way I could get the identity element, so what if $T$ is parametrized by both $\theta$ and $\phi$ ? I will surely get the identity element. Now how do I proceed from here?
 A: Your orthogonal matrix 
$$R(\phi,\theta)=
\begin{bmatrix}
    \cos(\theta)                &\sin(\theta)              & 0  \\
    \sin(\theta)\cos(\phi)       & -\cos(\theta)\cos(\phi)   & \sin(\phi)\\
    \sin(\theta)\sin(\phi)       & -\cos(\theta)\sin(\phi)   &-\cos(\phi)
\end{bmatrix}$$
must have antisymmetric generators.
To find them, you must expand around the origin, $\phi=\pi, \theta=0$. To allay confusion, define $\Phi\equiv = \pi-\phi$, so the origin is at $\Phi=\theta=0$.
$$R(\Phi,\theta)=
\begin{bmatrix}
    \cos(\theta)                &\sin(\theta)              & 0  \\
   - \sin(\theta)\cos(\Phi)       & \cos(\theta)\cos(\Phi)   & \sin(\Phi)\\
    \sin(\theta)\sin(\Phi)       & -\cos(\theta)\sin(\Phi)   &\cos(\Phi)
\end{bmatrix}$$
Evaluate    $R(\delta\Phi, 0)=\bigg[I+t\delta\Phi\bigg]$, so 
 $$t=
\begin{bmatrix}
    0                &0              & 0  \\
    0       & 0   & 1\\
    0      & - 1   &0
\end{bmatrix}.$$
Can you also evaluate $R(0,\delta \theta)$?

Note added as per comments.
The above rotation matrix R then, in the conventions of WP, is but 
$$
e^{-\Phi L_x} e^{-\theta L_z} ,
$$
which you might choose to compose by BCH, $$\exp (-\Phi L_x -\theta L_z+ \Phi  \theta [L_x,L_z]/2+... ), $$ or the Gibbs finite rotation formula, etc. if you were so inclined. For orthogonal axes like yours, Gibbs' formula all but collapses: the effective axis of rotation is just parallel to $\hat z \tan (\theta/2) +\hat x \tan (\Phi/2) +\hat y \tan (\theta/2) \tan (\Phi/2)$ ! (Can you see that this is precisely the invariant vector of R?)
In any case, the limiting procedure at the origin yielding the generators from your finite rotation matrix  has sacrificed information: convince yourself that several different rotation matrices may share this identical behavior at the origin, of course--think of reversing the order of the two factors above; so you should not expect to reconstitute this specific rotation matrix from the tangent space behavior at the origin, in general. (Here you already factored your finite rotations in advance. What Lie's 3rd theorem guarantees is essentially Euler's theorem: the two component rotations will combine to a single rotation about a new axis.) 
A: The parametrization you've given just isn't of the exponential form you're after. 
The matrix you've written down is parametrized by the location of the $x$ axis after the rotation, which has polar angle $\theta$ and azimuthal angle $\phi$ in polar spherical coordinates drawn around the old $x$ axis. It is not a rotation by angle $\phi$ about an axis at angle $\theta$, nor is it a rotation by angle $\theta$ about any clean axis, nor is it a combination of rotations by angles $\theta$ and $\phi$. It can of course be expressed as a rotation by a certain angle about a certain axis, but that angle is neither $\theta$ nor $\phi$.
As such, there is no useful way in which finding a low-order expansion in terms of $\theta$ or $\phi$ will give you a useful generator that will re-create your matrix upon exponentiation.
