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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.

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

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  • $\begingroup$ Good spot deserves a +1 $\endgroup$ – Kevin May 27 '16 at 7:36
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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.

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  • $\begingroup$ but $\sin x \cos x \ne 1$ for any x. After all, that would imply that $\sin 2x = 2$ $\endgroup$ – Doug M May 26 '16 at 20:41
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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 $$ \frac{\sin x}{\sin x+\cos x}=\frac{\sec x}{\sec x+\csc x} $$

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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.

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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).

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