How to solve this simple trignometric problem? So this is the question that was given in a textbook and i attempted to win from the book which was saying i was wrong?

If $$\frac{\sin\theta + \cos\theta}{\sin\theta  - \cos\theta} = \frac{5}{4}$$ then what is te value of$$\frac{\tan^2\theta + 1}{\tan^2\theta - 1}$$

so i attempted to solve it this way$$\implies\frac{\tan^2\theta + 1}{\tan^2\theta - 1} = \frac{\frac{\sin^2\theta}{\cos^2\theta} + 1}{\frac{\sin^2\theta}{\cos^2\theta} - 1}$$
$$=\frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta - \cos^2\theta}$$$$=\frac{1}{\sin^2\theta - \cos^2\theta}$$$$=\frac{1}{(\sin\theta + \cos\theta)(\sin\theta - \cos\theta)}$$substituting the values from$$=\frac{\sin\theta + \cos\theta}{\sin\theta  - \cos\theta} = \frac{5}{4}$$we get$$=\frac{1}{20}$$ but this is wrong answer so where am i wrong and what is the mistake
Note = I am not looking for the solution, i want to know what is wrong with this approach and why
Thanks 
Akash

Applying Componendo and dividendo on 
$$\dfrac{\sin\theta +\cos\theta}{\sin \theta -\cos\theta}=\dfrac54$$
to get $$\frac{\sin\theta}{\cos\theta}=\frac{5+4}{5-4}$$
$$\implies \frac{\tan^2\theta}1=\frac{9^2}1$$
Again apply Componendo and dividendo
