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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: $$\text{Left}=\frac{\sin^2{\theta}+\cos^2{\theta}+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{(\sin{\theta}+\cos{\theta})^2}{(\cos{\theta}+\sin{\theta})(\cos{\theta}-\sin{\theta})}=\frac{\sin{\theta}+\cos{\theta}}{\cos{\theta}-\sin{\theta}} \\\text{Right}=\frac{1+\frac{\sin{\theta}}{\cos{\theta}}}{1-\frac{\sin{\theta}}{\cos{\theta}}}=\frac{\cos{\theta}+\sin{\theta}}{\cos{\theta}-\sin{\theta}} \\{\therefore}\text{Left=Right}$$

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    $\begingroup$ In your attempt, you have not expanded the brackets correctly, and your cancelling is incorrect: $-\cos^2\theta+\sin^2\theta$ does not cancel with $1$ (whereas $-\cos^2\theta-\sin^2\theta$ would cancel with $1$) $\endgroup$ – rbird Aug 19 '18 at 13:48
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    $\begingroup$ Why do you cancelled out $1-\cos^2(\theta)+\sin^2(\theta)$? Seems this is where you were wrong [assume you've expanded right]. $\endgroup$ – xbh Aug 19 '18 at 13:49
  • $\begingroup$ Thank you both, I think that's where I was wrong. $\endgroup$ – student Aug 19 '18 at 14:02
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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}

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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.

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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}

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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:$$\frac{1+2sc}{c^2-s^2}=\frac{1+t}{1-t}$$

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  • $\begingroup$ @student ,what you did with LHS and the result is correct, no need any operation on RHS.. $\endgroup$ – sirous Aug 19 '18 at 17:30
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$\dfrac{1+\tan t}{1-\tan t} =$

$\dfrac{\cos t + \sin t}{\cos t -\sin t}=$

$\dfrac{(\cos t+\sin t)^2}{\cos^2 t - \sin^2t}=$

$\dfrac{1+2\sin t \cos t}{\cos^2 t-\sin^2 t}$.

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