Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer
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 $\sec(A) \cos(A) = 1$ : $$ \frac{\sec(A)}{\sec(A) + \csc(A) } $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.