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how to prove $\tan A+\sec A=\frac{1}{(\sec A-\tan A)}$ ?

I already tried: $$\begin{align} \sin A/\cos A+1/\cos A&=1/(\sec A-\tan A)\\ \sin A+1/\cos A&=1/(\sec A-\tan A)\\ \end{align}$$

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  • $\begingroup$ I never understood why people post solutions when someone has almost said isomething identitcal before them... I am somewhat glad these answers didn't get undeserved reputations because everyone understand the answers... $\endgroup$ – Lost1 Jan 7 '14 at 22:26
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    $\begingroup$ @Lost1 Several answers were posted within ~1-2 minutes of each other. Myself, I never understood why so many on this site are obsessed with exterminating duplication... $\endgroup$ – Slade Jan 7 '14 at 22:32
  • $\begingroup$ @User-33433 no they are not... Id stop writing when i see a duplicate answer... At the time i write this comment, they were written 48mins - 1 hr plus ago. I do not believe all of them start writing before seeing at least 1 answer has been posted... $\endgroup$ – Lost1 Jan 7 '14 at 23:05
  • $\begingroup$ @Lost1 which answers you're referring to and which of them do you think appeared first? $\endgroup$ – Kaster Jan 8 '14 at 1:52
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$\mathbf{HINT:}$$$ \begin{align} (\tan A+\sec A)(\sec A-\tan A)&=\require{cancel}\cancel{\tan A\sec A}-\tan^2 A+\sec^2 A\,\, \require{cancel}\cancel{\ -\tan A \sec A} \\ &=\sec^2 A -\tan^2 A\equiv1 \end{align}$$

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Hint: $\tan^2 \alpha + 1 = \sec^2 \alpha$.

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Sometimes we just need to stop the sin/cos madness. In many situations it is certainly appropriate to convert everything into sines and cosines, but in this problem it just adds unnecessary complexity.

First, analyze the problem as follows

$$\tan A+\sec A = \frac{1}{\sec A - \tan A}$$ $$(\tan A+\sec A)(\sec A - \tan A) = (\frac{1}{\sec A - \tan A})(\sec A - \tan A)$$ $$\sec^2A-\tan^2A = 1$$

But this is just a rearrangment of the familiar identity

$$\sec^2A=\tan^2A+1$$ The preceding suggests that we need to multiply top and bottom by $\sec A-\tan A$ $$\tan A+\sec A=(\tan A+\sec A)\frac{\sec A-\tan A}{\sec A-\tan A}=\dots$$

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$$\sin A+1/\cos A=1/(\sec A-\tan A)$$

Parentheses! And let's remember that we haven't shown this yet.

$$(\sin A+1)/\cos A=^? 1/(\sec A-\tan A)$$

Ahh, much better. Well, why stop there?

$$(\sin A+1/\cos A)=^? 1/((1/\cos A)-(\sin A / \cos A))$$

Well, get simplifying. It's impossible to fail to solve a problem; it's only possible to stop trying.

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HINT: $$\tan A+\sec A=\sin A/\cos A+1/\cos A=\frac{\sin A+1}{\cos A}\frac{1-\sin A}{1-\sin A}=\ldots$$ $$\ldots=\frac{\cos A}{1-\sin A}=\frac{\cos A/\cos A}{1/\cos A-\sin A/\cos A}$$

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See $\tan A$ as $\frac{\ sinA}{\cos A}$ and $\sec A$ as $\frac{1}{\cos A}$

1 part --------> $$\frac{\sin A+1}{ \cos A}$$ , 2 part---------> $$\frac{\cos A}{1-\sin A}$$

Focuse on the second part, multiply by $$\frac{1+\sin A}{1+\sin A}$$

So

$$\frac{\cos A}{1-\sin A}*\frac{1+\sin A}{1+\sin A}$$

Remembering that $\cos^2A=1-\sin^2A$ you will obtain

$$\frac{\sin A+1}{\cos A}=\frac{\sin A+1}{\cos A}$$

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$\sec ^2\alpha -\tan ^2\alpha=1$

$(\sec \alpha + \tan \alpha) (\sec \alpha - \tan \alpha)=1$

$\implies \sec \alpha + \tan \alpha =\frac{1}{\sec \alpha - \tan \alpha}$

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