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 ).
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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Divide numerator and denominator by $\ds{\cos^{2}\pars{\alpha}=1/\sec^{2}\pars{\alpha}}$:

\begin{align}
&\color{#66f}{\large{1 - 2\sin\pars{\alpha}\cos\pars{\alpha} \over 1 - 2\sin^{2}\pars{\alpha}}}
={\sec^{2}\pars{\alpha} - 2\sin\pars{\alpha}/\cos\pars{\alpha}\over
\sec^{2}\pars{\alpha} - 2\sin^{2}\pars{\alpha}/\cos^{2}\pars{\alpha}}
\end{align}

With identities $\ds{\sec^{2}\pars{\alpha}=\tan^{2}\pars{\alpha} + 1}$ and
$\ds{\tan\pars{\alpha}=\sin\pars{\alpha}/\cos\pars{\alpha}}$:

\begin{align}
&\color{#66f}{\large{1 - 2\sin\pars{\alpha}\cos\pars{\alpha} \over 1 - 2\sin^{2}\pars{\alpha}}}
={\tan^{2}\pars{\alpha} + 1 - 2\tan\pars{\alpha}\over
\tan^{2}\pars{\alpha} + 1 - 2\tan^{2}\pars{\alpha}}
={\bracks{1 - \tan\pars{\alpha}}^{2} \over 1 - \tan^{2}\pars{\alpha}}
\\[5mm]&={\bracks{1 - \tan\pars{\alpha}}^{2}\over
\bracks{1 - \tan\pars{\alpha}}\bracks{1 + \tan\pars{\alpha}}}
={1 - \tan\pars{\alpha} \over 1 + \tan\pars{\alpha}}={1 - 3/4\over 1 + 3/4}
=\color{#66f}{\large{1 \over 7}} \approx {\tt 0.1429}
\end{align}

A: converting $\tan(x)$ in $\tan(\alpha)=\frac{2\tan(\alpha/2)}{1-\tan(\alpha/2)}$
solving now the equation
$\frac{2\tan(\alpha/2)}{1-\tan(\alpha/2)}=3/4$ for $\tan(\alpha/2)$ we get
$\tan(\alpha/2)=-3$ or $1/3$
now use that 
$\sin(\alpha)=2\,{\frac {\tan \left( \alpha/2 \right) }{1+ \left( \tan \left( \alpha
/2 \right)  \right) ^{2}}}$
and 
$\cos(\alpha)={\frac {1- \left( \tan \left( \alpha/2 \right)  \right) ^{2}}{1+
 \left( \tan \left( \alpha/2 \right)  \right) ^{2}}}
$
and you can solve your problem
A: The fact that $\tan\alpha=\frac34$ in a right angled triangle implies that the opposite side is $3$ and the adjacent side is $4$. Using the Pythagorean Theorem, the hypotenuse is $\sqrt{3^2+4^2}=5$.
Therefore, $\sin\alpha=\frac{3}{5}$ and $\cos\alpha=\frac{4}{5}$.
Simply plug this into your expression to find its value.
A: 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.
