<s>[INST] 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
 [/INST] Observe that $\dfrac a b=\dfrac c d$ doesn't imply $a=c$ and $b=d$.
Now you may proceed as follows,
\begin{align}
&\dfrac{\sin \theta +\cos\theta}{\sin \theta -\cos\theta}=\dfrac54
\\\implies&4\sin \theta +4\cos\theta=5\sin \theta-5\cos\theta
\\\implies&\sin\theta=9\cos\theta.
\end{align}
</s>