# A trigonometric proof

How to prove the following? $${1-\sin A \over1+\sin A} = (\sec A- \tan A)^2$$

this is what I've done till now: \begin{array}{ccc} {1-\sin A \over1+\sin A} &=& {1+\sin^2 A - 2\sin A \over 1-\sin^2A} \\ \\ &=&1+1-\cos^2A-2\sin A \over \cos^2A \end{array}

How shall I proceed?

EDIT: I've got my answer, but I now have another question. Can this be proved by working on the Right hand side of the equation? In this case, how would you infer the LHS from RHS?

-
you are already done just use that $1-cos^2A=sinA$. do you see the solution now? – Albanian_EAGLE Apr 22 '13 at 17:48
@Albanian_EAGLE You're missing a square in $\sin ^2 A$. OP is close, but not immediately there yet. – Calvin Lin Apr 22 '13 at 17:51
true, I am sorry. how can I edit my comment? – Albanian_EAGLE Apr 22 '13 at 17:54
@Albanian_EAGLE Now, you can't. The editing possibility expires after 3 min, I think. – Américo Tavares Apr 22 '13 at 17:57

## 2 Answers

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

-
I think you mean "$\equiv$" in place of "$=$". Writing $f(x)=g(x)$ asks the question: for which values of $x$ do $f$ and $g$ agree? Writing $f(x) \equiv g(x)$ makes the statement that $f$ and $g$ agree for all values of $x$. – Fly by Night Apr 22 '13 at 17:55
@FlybyNight No one asked question so far. – Kaster Apr 22 '13 at 22:18

As a matter of strategy, I would advise working with the right-hand side instead. The reason is non-mathematical: we are much more familiar with $\sin$ and $\cos$ than with $\sec$, and perhaps even $\tan$.

Rewrite $\sec A$ and $\tan A$ in terms of sine and cosine. We get $$\left(\frac{1}{\cos A}-\frac{\sin A}{\cos A}\right)^2=\frac{(1-\sin A)^2}{\cos^2 A}.$$ Now, if we take a little peek at the left hand side, it is natural to replace $\cos^2 A$ by $1-\sin^2 A$, and note that this factors as $(1-\sin A)(1+\cos A)$.

-
I think you mean "$\equiv$" in place of "$=$". Writing $f(x)=g(x)$ asks the question: for which values of $x$ do $f$ and $g$ agree? Writing $f(x) \equiv g(x)$ makes the statement that $f$ and $g$ agree for all values of $x$. – Fly by Night Apr 22 '13 at 17:59
@FlybyNight: That would be a matter of "local" convention. The symbol $=$ is also used to mean identically equal, I would guess more frequently than $\equiv$. In principle, it might be nice if notation avoided all ambiguities. In practice, it seems to be useful to "overload" certain symbols. – André Nicolas Apr 22 '13 at 18:17
The addendum "for all $x$" should be used when "=" is being used to denote an identity. Otherwise, how do we distinguish between "$x=1$" and "$\cos^2x + \sin^2x = 1$"? – Fly by Night Apr 22 '13 at 18:25
en.wikipedia.org/wiki/… – Fly by Night Apr 22 '13 at 18:26
@AméricoTavares: An interesting feature of school trigonometric identities is that they are purely formal. For example, in the identity of the OP, the left side is defined at $\pi/2$, and the right side is not, but one is not expected even to comment. – André Nicolas Apr 22 '13 at 21:13