Can anyone offer please help me solve the following trig identity.
$$\frac{1-\tan(x/2)}{1+\tan(x/2)}=\frac{1-\sin x}{\cos x}=\frac{\cos x}{1+\sin x}$$
My work thus far has been on the left most side.
I did
$$\frac{\;\;1-\dfrac{\sin(x/2)}{\cos(x/2)}\;\;}{1+\dfrac{\sin(x/2)}{\cos(x/2)}}$$
Unfortunately now I am not sure what to do I multiplied the numerator by $\cos$ on the the top and $\cos^2$ on the bottom to get $\cos x$,but I am not sure if this is correct.
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$\begingroup$In order to make typing easier, I will let $t=x/2$. Replacing $\tan t$ by $\frac{\sin t}{\cos t}$, as you did, is a good general purpose strategy. So the leftmost side is equal to $$\frac{1-\frac{\sin t}{\cos t}}{1+\frac{\sin t}{\cos t}}.$$ The next step is semi-automatic. It seems sensible to bring top and bottom to the common denominator $\cos t$, and "cancel." We get $$\frac{\cos t -\sin t}{\cos t+\sin t}.$$ The move after that is not obvious: Multiply top and bottom by $\cos t-\sin t$.
At the bottom we get $\cos^2 t-\sin^2 t$, which from a known double-angle formula we recognize as $\cos 2t$.
The top is $(\cos t-\sin t)^2$. Expand. We get $\cos^2 t-2\sin t\cos t+\sin^2 t$. The $\cos^2 t+\sin^2 t$ part is equal to $1$, and the $2\sin t\cos t$ part is equal to $\sin 2t$. So putting things together we find that the leftmost side is equal to $$\frac{1-\sin 2t}{\cos 2t},$$ which is exactly what we want.
The second identity $\frac{1-\sin x}{\cos x}=\frac{\cos x}{1+\sin x}$ is much easier. In $\frac{1-\sin x}{\cos x}$, multiply top and bottom by $1+\sin x$. On top we now get $1-\sin^2 x$, which is $\cos^2 x$. At the bottom we get $\cos x(1+\sin x)$. Cancel a $\cos x$.
Remark: We can play a game of making the calculation more magic-seeming. The middle expression is equal to $\frac{1+\sin 2t}{\cos 2t}$. Replace the $1$ on top by $\cos^2 t+\sin^2 t$, and the $\sin t$ by $2\cos t\sin t$. Then on top we have $(\cos t+\sin t)^2$. Replace the $\cos 2t$ at the bottom by $\cos^2 t-\sin^2 t$, which is $(\cos t+\sin t)(\cos t -\sin t)$. Cancel the $\cos t+\sin t$, and we arrive at $$\frac{\cos t+\sin t}{\cos t-\sin t}.$$ Now divide top and bottom by $\cos t$, and we get the desired $\frac{1+\tan t}{1-tan t}$.
$\endgroup$ 2 $\begingroup$I gave a proof without words of the identity $$ \cot(\theta/2)=\frac{\sin(\theta)}{1-\cos(\theta)} $$ which is the same as $$ \tan(\theta/2)=\frac{1-\cos(\theta)}{\sin(\theta)} $$ writing $\theta=\pi/2-x$, we get $$ \frac{1-\tan(x/2)}{1+\tan(x/2)}=\tan\left(\pi/4-x/2\right)=\frac{1-\sin(x)}{\cos(x)} $$
$\endgroup$ $\begingroup$proof: $$ \frac{1-\tan(x/2)}{1+\tan(x/2)}=\frac{1-\frac{\sin(x/2)}{\cos(x/2)}}{1+\frac{\sin(x/2)}{\cos(x/2)}}=\frac{\cos(x/2)-\sin(x/2)}{\cos(x/2)+\sin(x/2)} $$ Now multiply numerator and denomenetor by $\cos(x/2)-\sin (x/2)$, then we get $$ \frac{\cos^2(x/2)+\sin^2(x/2)-2\sin (x/2)\cos(x/2)}{\cos^2(x/2)-\sin^2(x/2)} =\frac{1-\sin x}{\cos x} $$ We know that $\sin x=2\sin(x/2)\cos(x/2)$ and $\cos x=\cos^2(x/2)-\sin^2(x/2)$
Thus we have required result.
$\endgroup$ $\begingroup$I'm doing it from RHS
$$\frac{1-\tan(x/2)}{1+\tan(x/2)}=\frac{1-\sin x}{\cos x}=\frac{\cos x}{1+\sin x}$$
$$\frac{\cos x}{1+\sin x}$$
$$ \frac {\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}} {1+\frac {2tan(x/2)}{1+\tan^2(x/2)}} $$
$$ \frac {\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}} {\frac {1+tan^2(x/2)+2tan(x/2)}{1+\tan^2(x/2)}} $$
$$ \frac {1-\tan^2(x/2)}{1+tan^2(x/2)+2tan(x/2)} $$
$$ \frac {(1-\tan(x/2))(1+\tan(x/2))}{(1+tan(x/2))^2} $$
$$ \frac {(1-\tan(x/2))}{(1+\tan(x/2))} $$ =LHS
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