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$$\lim _{ \theta \rightarrow \frac { \pi }{ 4 } }{ \frac { 1 }{\tan\theta -1 } -\frac { 2 }{ \tan {^ 2\theta -1 } } } $$

Steps I took:

$$\begin{align}\lim _{ \theta \rightarrow \frac { \pi }{ 4 } }{ (\frac { 1 }{ \tan\theta -1 } -\frac { 2 }{ \sec^{ 2 }\theta } ) } &=(\frac { 1 }{ \frac { \sin\theta }{ \cos\theta } -\frac { \cos\theta }{ \cos\theta } } -\frac { 2 }{ \frac { 1 }{ \cos^{ 2 }\theta } } )\\&=\frac { 1 }{ \frac { \sin\theta -\cos\theta }{ \cos\theta } } -\frac { 2\cos^{ 2 }\theta }{ 1 } \\&=\frac { \cos\theta }{ \sin\theta -cos\theta } -\frac { 2\cos^{ 2 }\theta }{ 1 } \\&=\frac { \cos\theta }{ \sin\theta -\cos\theta } -\frac { 2\cos^{ 2 }\theta }{ 1 } \cdot \frac { (\sin\theta -\cos\theta ) }{ (\sin\theta -\cos\theta ) } \\&=\frac { \cos\theta -2\cos^{ 2 }\theta (\sin\theta -\cos\theta ) }{ \sin\theta -\cos\theta } \\&=\end{align}$$

I am now stuck and see no way to go any further.

Please provide a hint so that I can arrive at the correct solution by myself (hopefully).

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    $\begingroup$ Write your expression with a common denominator of $\tan^2\theta - 1$. You'll find it becomes rather nice. $\endgroup$ – Simon S Jan 14 '15 at 19:35
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    $\begingroup$ Lots of spoon-feeding going on in these answers! And yet nobody took the time to demonstrate the manipulation of the L.H.S... $\endgroup$ – The Chaz 2.0 Jan 14 '15 at 19:42
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Your starting step is where you went wrong: $$ \tan^2 \theta -1 \neq \sec^2 \theta \\ \tan^2 \theta +1 = \sec^2 \theta $$

Note that $$\frac{1}{\tan \theta -1} - \frac{2}{\tan^2 \theta -1} = \frac{\tan \theta +1-2}{\tan^2 \theta -1} =\frac{\tan \theta -1}{\tan^2 \theta -1} = \frac{1}{\tan \theta +1} $$

How does this behave when $\tan \theta$ becomes large?

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  • $\begingroup$ I should have realized that the denominator on the RHS is a difference of squares... I am going to work with this new realization and see where it gets me. I'll let you know $\endgroup$ – Cherry_Developer Jan 14 '15 at 19:55
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Try using the equality

$$\frac{1}{\tan\theta -1}-\frac{2}{\tan^2\theta -1}=\frac{1}{1+\tan\theta}$$

and go from there.

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it is equivalent to $\frac{\cos(\theta)}{\sin(\theta)+\cos(\theta)}$

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After adding these two fractions you will notice that a lot of things reduce. $$\frac{1}{\tan \theta - 1} - \frac{2}{\tan^2 \theta - 1} = \frac{\tan \theta + 1 - 2}{\tan^2 \theta - 1} = \frac{1}{\tan \theta + 1}$$

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$\lim _{ \theta \rightarrow \frac { \pi }{ 4 } }{ (\frac { 1 }{ tan\theta -1 } -\frac { 2 }{ sec^{ 2 }\theta } )} = \lim_{\theta \rightarrow \frac {\pi}{4}} \frac{1}{\tan\theta+1} = \frac12$

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$$\begin{align}\lim_{\theta \to \frac{\pi}{4}} \frac{(\tan \theta + 1)}{(\tan \theta -1)(\tan \theta + 1)} - \frac{2}{\tan^2\theta - 1} &= \lim_{\theta \to \frac{\pi}{4}} \frac{\tan \theta + 1 - 2}{\tan^2\theta - 1} \\&= \lim_{\theta \to \frac{\pi}{4}} \frac{\tan \theta - 1}{\tan^2\theta - 1} \\&= \lim_{\theta \to \frac{\pi}{4}}\frac{1}{\tan \theta + 1}\end{align}$$

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