<s>[INST] Find the exact value of $\frac{1-2\sin\left(α\right)\cos\left(α\right)}{1-2\sin^2\left(α\right)}$. The task is: Given that
$0\leq \alpha \leq \pi$ is an angle with $\tan\left(\,\alpha\,\right)=3/4$, find the exact value of
$$
{1 - 2\sin\left(\,\alpha\,\right)\cos\left(\,\alpha\,\right)\over
 1 - 2\sin^{2}\left(\,\alpha\,\right)}.
$$
I have stuck with this task. Can anybody help me with this ?
( I need detailed answer ).
 [/INST] Although the OP has already accepted an answer, I'll provide a different approach, even because, together with the other, it did come to my mind earlier, but I had to leave the computer.
$$\displaystyle \frac{1-2\sin\alpha \cos\alpha}{1-2\sin^2 \alpha}=\frac{1-2\sin\alpha\cos\alpha}{\cos^2\alpha-\sin^2\alpha},$$ which, recalling the double-angle formulae, becomes $$\displaystyle \frac{1-\sin2\alpha}{\cos2\alpha}.$$
Finally use $\displaystyle \sin\alpha=\frac{2\tan\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}$ and $\displaystyle \cos\alpha=\frac{1-\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}$ to obtain $$\displaystyle  \frac{1-\frac{2\tan\alpha}{1+\tan^2\alpha}}{\frac{1-\tan^2\alpha}{1+\tan^2\alpha}}=\frac{1-\frac{3}{2}\cdot\frac{16}{25}}{\frac{7}{16}\cdot \frac{16}{25}}=\frac{1}{25}\cdot\frac{25}{7}=\frac{1}{7}.$$
Edit: I notice Dr. Sonnhard had written something similar, but I guess this is more expanded.
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