# Verify trigonometry equation $\frac{\sin(A)}{\sin(A) + \cos(A)}=\frac{\sec(A)}{\sec(A)+\cos(A)}$

How would I verify the following trig equation? $$\frac{\sin(A)}{\sin(A) + \cos(A)}=\frac{\sec(A)}{\sec(A)+\cos(A)}$$

My work so far is to write the RHS as $$\frac{1/\cos(A)}{1/\cos(A) + \cos(A)}$$

But I am not sure what I can do to prove the identity.

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What you have is incorrect. For instance, take $A=0$, the LHS evaluates to $0$ while the RHS gives us $\dfrac12$. – user17762 Jul 15 '12 at 22:59
There is nothing you can do to prove the identity, because it is not a identity. – copper.hat Jul 15 '12 at 23:02
Maybe the $\cos$ on the RHS of the identity should be $\csc$? – TMM Jul 15 '12 at 23:03
I wrote It out of the book I am using. – Fernando Martinez Jul 15 '12 at 23:04
Well the typo is not on my part but on the book I am using. – Fernando Martinez Jul 15 '12 at 23:08

If $\frac{a}{a+x} = \frac{b}{b+x}$, and you have that $x\neq 0$ (and the denominators too), then multiplying across and canceling will give $a=b$.

So, the equation is satisfied only if $\sin A = \frac{1}{\cos A}$, which is impossible.

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so would I cross multiply? – Fernando Martinez Jul 15 '12 at 23:05
You can do what you want, but the formula as stated will never be true. – copper.hat Jul 15 '12 at 23:08

I assume there is a typo: $$\dfrac{\sin(A)}{\sin(A) + \cos(A)}=\dfrac{\sec(A)}{\sec(A)+\csc(A)}$$

Divide numerator & denominator by ${\sin(A)},$ and use that $\frac{1}{\sin(A)} = \csc(A)$:

$$\frac{\sin(A)}{\sin(A) + \cos(A)} = \frac{1}{1 + \frac{1}{\sin(A)} \cos(A)} = \frac{1}{1 + \csc(A) \cos(A)}$$

Now, multiply numerator & denominator by by $\sec(A),$ and use the fact $\sec(A) \cos(A) = 1$ : $$\frac{\sec(A)}{\sec(A) + \csc(A) }$$

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