Prove that $\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$ Prove that $$\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$$
Here's my attempt
$$\require{cancel}\text{Left - Right} = \frac{(1+2\sin{\theta}\cos{\theta})(1-\tan{\theta})-(1+\tan{\theta})(\cos^2{\theta}-\sin^2{\theta})}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}=
\frac{\cancel1\cancel{-\tan{\theta}}+2\sin^2{\theta}\cancel{+2\sin{\theta}\cos{\theta}}\cancel{-\cos^2{\theta}+\sin^2{\theta}}\cancel{-\sin{\theta}\cos{\theta}}\cancel{+\tan{\theta}}\cancel{-\sin{\theta}\cos{\theta}}}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}
=\frac{2\sin^2{\theta}}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}$$
It should be zero, but it isn't? Here's the proof:
$$\text{Left}=\frac{\sin^2{\theta}+\cos^2{\theta}+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{(\sin{\theta}+\cos{\theta})^2}{(\cos{\theta}+\sin{\theta})(\cos{\theta}-\sin{\theta})}=\frac{\sin{\theta}+\cos{\theta}}{\cos{\theta}-\sin{\theta}}
\\\text{Right}=\frac{1+\frac{\sin{\theta}}{\cos{\theta}}}{1-\frac{\sin{\theta}}{\cos{\theta}}}=\frac{\cos{\theta}+\sin{\theta}}{\cos{\theta}-\sin{\theta}}
\\{\therefore}\text{Left=Right}$$
 A: Direct way:
Factor out $\cos^2\theta$ from numerator and denominator of the l.h.s. fraction and do some trigonometry:
$$\frac{1+2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}=\frac{\dfrac1{\cos^2\theta}+2\tan\theta}{1-\tan^2\theta}=\frac{1+\tan^2\theta+2\tan\theta}{1-\tan^2\theta}=\frac{(1+\tan\theta)^2}{1-\tan^2\theta}$$
and simplify.
A: Note that
$$
1+2\sin\theta\cos\theta=\cos^2\theta+\sin^2\theta+2\sin\theta\cos\theta=
(\cos\theta+\sin\theta)^2
$$
Therefore
\begin{align}
\frac{1+2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}
&=\frac{(\cos\theta+\sin\theta)^2}{(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)}
\\[6px]
&=\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}
\\[6px]
&=\frac{\cos\theta(1+\tan\theta)}{\cos\theta(1-\tan\theta)}
\\[6px]
&=\frac{1+\tan\theta}{1-\tan\theta}
\end{align}
A: Use 
\begin{equation}
 2 \sin \theta \cos \theta
 =
 \sin 2\theta 
 =
 \frac{2 \tan \theta}{1 + \tan^2 \theta}
\end{equation}
\begin{equation}
\cos^2 \theta - \sin^2 \theta
=
 \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}
\end{equation}
Replacing the above quantities, we get
\begin{equation}
 \frac{1 + \frac{2 \tan \theta}{1 + \tan^2 \theta}}{\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}}
 =
 \frac{1 + \tan^2 \theta + 2 \tan \theta}{1 - \tan^2 \theta}
 =
 \frac{(1 + \tan \theta)^2}{( 1 - \tan \theta) (1 + \tan \theta)}
 =
 \frac{1 + \tan \theta}{1 - \tan \theta}
\end{equation}
A: Abbreviating $\sin\theta$ by $s$ and $\cos\theta$ by $s$ we have on base of $c^2+s^2=1$:$$1+2sc=c^2+2sc+s^2=(c+s)^2$$ so that:$$(1+2sc)(c-s)=(c+s)^2(c-s)=(c^2-s^2)(c+s)$$
Dividing boths sides by $c$ and abbreviating $\tan(\theta)$ by $t$ we get on base of $t=s/c$:$$(1+2sc)(1-t)=(c^2-s^2)(1+t)$$
Now divide both sides by $(1-t)(c^2-s^2)$ to achieve:$$\frac{1+2sc}{c^2-s^2}=\frac{1+t}{1-t}$$
A: $\dfrac{1+\tan t}{1-\tan t} =$
$\dfrac{\cos t + \sin t}{\cos t -\sin t}=$
$\dfrac{(\cos t+\sin t)^2}{\cos^2 t - \sin^2t}=$
$\dfrac{1+2\sin t \cos t}{\cos^2 t-\sin^2 t}$.
