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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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    it is in schaums outlines of trigonometry page 93 problem 8.382012-07-15

2 Answers 2

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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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    You can do what you want, but the formula as stated will never be true.2012-07-15
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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) } $