# How would you verify the following trig identity $\frac{\sin(x)}{\sin(x)+\cos(x)}=\frac{\sec(x)}{\sec(x)+\cos(x)}$

How to verify the following trig identity.

$$\frac{\sin(x)}{\sin(x)+\cos(x)}=\frac{\sec(x)}{\sec(x)+\cos(x)}$$

I started with the right side and multiplied the numerator and denominator by $\sec(x)-\cos(x)$

then I got

$$\frac{\sec^2(x)-1}{\sec^2(x)-\cos^2(x)}=\frac{\tan^2(x)}{\sec^2(x)-\cos^2(x)}$$

but now I am stuck.

• The identity is false: set $x=0$ and the left hand side is $0$, but the right hand side is $1/2$. – egreg May 26 '16 at 20:35
• Your identity isn't an identity. So pick a value of x and show that it is not true. – Doug M May 26 '16 at 20:35
• Oh I guess my book must have a typo or something fo sure. – Fernando Martinez May 26 '16 at 20:37
• If your identity was true it would imply that $\sin x = \sec x$ for all real $x$. – Jeevan Devaranjan May 26 '16 at 20:37

There may be a typo, either in your question or in your book. Is it possible it was meant to be $$\frac{\sin(x)}{\sin(x)+\cos(x)}=\frac{\sec(x)}{\sec(x)+\csc(x)}$$ instead (note "$\csc$" instead of "$\cos$")?

If so, you can get this by just dividing each term on the left by $\sin x \cos x$.

• Good spot deserves a +1 – Kevin May 27 '16 at 7:36

If you could cross multiply this, you will get:

$sin(x)(sec(x)+cos(x))=sec(x)(sin(x)+cos(x))$

$\Rightarrow sin(x)(\frac{1}{cos(x)}+cos(x))=\frac{1}{cos(x)}(sin(x)+cos(x))$

$\Rightarrow \frac{sin(x)}{cos(x)}+sin(x)cos(x)=\frac{sin(x)}{cos(x)}+\frac{cos(x)}{cos(x)}$

$\Rightarrow sin(x)cos(x)=1$

$\Rightarrow 2sin(x)cos(x)=sin(2x)=2$

There's no such $x$ such that it holds. So it is not a valid identity.

• but $\sin x \cos x \ne 1$ for any x. After all, that would imply that $\sin 2x = 2$ – Doug M May 26 '16 at 20:41

It's easy to see that the right hand side is $(1+\cos^2x)^{-1}$; so the equality holds if and only if $$\sin x(1+\cos^2x)=\sin x+\cos x$$ that is $$\sin x\cos^2x-\cos x=0$$ Since $\cos x\ne0$ as you need to use the secant, it only remains $\sin x\cos x=1$, which has no solution.

So the equality holds for no value of $x$.

On the other hand $$\frac{\sec x}{\sec x+\csc x}= \frac{\dfrac{1}{\cos x}}{\dfrac{1}{\cos x}+\dfrac{1}{\sin x}}= \frac{\sin x}{\sin x+\cos x}$$ So the correct identity is $$\frac{\sin x}{\sin x+\cos x}=\frac{\sec x}{\sec x+\csc x}$$

If both sides are defined (i.e. $\;x\neq \dfrac\pi2, \dfrac{3\pi}4\mod\pi$) \begin{align*}\frac{\sin x}{\sin x+\cos x}&=\frac{\sec x}{\sec x+\cos x}\iff \sin x(\sec x+\cos x)=(\sin x+\cos x)\sec x\\ &\iff \sin x \cos x=\sec x\cos x\iff\sin x =\sec x\iff\frac12\sin2x=1,\\ \end{align*} which has no solution.

This is not an identity. Because,when we putting the value of both sides of the identity we get two different value. Left hand value is 0.Right hand value is (1÷2).