# How to find $\cos A$ from $\sec$ and $\tan$?

It's given that
$$\sec A + \tan A = 4$$ How would you find $\cos A$ from this?

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### A simple solution!

$\sec^2A=(4-\tan A)^2$

So, you get $\tan A=\frac{15}8$ , by using $\sec^A=1+\tan^2A$

Now solve for $\cos A$.

### Or a big one.

$\sec A+\tan A=\dfrac{1+\sin A}{\cos A}=4.$

$(1+\sin A)^2=(4\cos A)^2$

Squaring both sides .

$17\cos^2A-2=2\sin A$

Now again square and you get a bi-quadratic in $\cos A$, by using $\sin^2A=1-\cos^2A$ as done above.

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Thanks for giving more than one solution! – Ghost Apr 19 '13 at 9:25
@exploringnet, we should try to avoid squaring which leaves the burden to eradicate extraneous roots – lab bhattacharjee Apr 19 '13 at 11:02
@labbhattacharjee But this don't have one of that kind.$\cos A=0$ is not a solution. – ABC Apr 19 '13 at 11:12
@exploringnet, I was talking about the "big one". What are its roots? – lab bhattacharjee Apr 19 '13 at 11:27
@labbhattacharjee Don't know, i certainly didn't solved it to the end.Is it giving any problem? – ABC Apr 19 '13 at 13:04

As $\sec^2A-\tan^2A=1,$

$$\sec A+\tan A=4\iff \sec A-\tan A=\frac14$$

Adding we get $$2\sec A=4+\frac14=\frac{17}4 .$$

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$\sec A=\frac{1}{\cos A}$ and $\tan A = \frac{\sin A}{\cos A}$.

Then you can later use $\sin^2 A =1-\cos^2A$.

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