Prove the trigonometric identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$ While solving an equation i came up with the identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$.
Prove whether this is really true or not. I can add that $$\tan\left(\frac{x}{2}\right) = \sqrt{\frac{1-\cos x}{1+\cos x}}$$
 A: The identity is true.
\begin{align*}
\cos x + \sin x \tan \frac{x}{2} & =(\cos^2\frac{x}{2}-\sin^2\frac{x}{2})+(2\sin \frac{x}{2}\cos \frac{x}{2})(\frac{\sin \frac{x}{2}}{\cos\frac{x}{2}})\\
& = \cos^2\frac{x}{2}+\sin^2 \frac{x}{2}\\
& = 1
\end{align*}
A: $$\sin2y\cdot\tan y=2\sin y\cos y\cdot\dfrac{\sin y}{\cos y}=1-\cos2y$$
OR
$$\dfrac{1-\cos2y}{\sin2y}=\dfrac{2\sin^2y}{2\sin y\cos y}=\tan y$$
A: Yup! We have $$\cos x + \sin x \tan \frac{x}{2} = (\cos ^2 \frac{x}{2} - \sin ^2 \frac{x}{2}) + (2 \sin \frac{x}{2} \cos \frac{x}{2}) \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \cos ^2 \frac{x}{2} + \sin ^2 \frac{x}{2} = 1$$
A: if $\tan\frac x2=t,$ then $\cos x =\frac{1-t^2}{1+t^2}, \sin x=\frac{2t}{1+t^2}$ so it works!
A: Let me try. You have: $$\cos x + \sin x \tan (\frac{x}{2}) = 1 - 2\sin^2 (\frac{x}{2}) + 2 \sin (\frac{x}{2}) \cos (\frac{x}{2}) \tan (\frac{x}{2}) = 1 -2\sin^2 (\frac{x}{2}) + 2\sin^2 (\frac{x}{2}) = 1.$$
A: Notice, $$LHS=\cos x+\sin x\tan \frac{x}{2}$$
$$=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}+\frac{2\tan \frac{x}{2}}{1+\tan^2\frac{x}{2}}\tan \frac{x}{2}$$
$$=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}+\frac{2\tan^2 \frac{x}{2}}{1+\tan^2\frac{x}{2}}$$
$$=\frac{1-\tan^2\frac{x}{2}+2\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$
$$=\frac{1+\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}=1=RHS$$
A: We have $$\cos\left(x\right)+\sin\left(x\right)\tan\left(\frac{x}{2}\right)=1\Leftrightarrow\cos\left(x\right)\cos\left(\frac{x}{2}\right)+\sin\left(x\right)\sin\left(\frac{x}{2}\right)=\cos\left(\frac{x}{2}\right)
 $$ and, using product to sum identity and the fact that cos is a even function$$\cos\left(x\right)\cos\left(\frac{x}{2}\right)=\frac{1}{2}\cos\left(\frac{x}{2}\right)+\frac{1}{2}\cos\left(\frac{3x}{2}\right)
 $$ and again using product to sum identity we get $$\sin\left(x\right)\sin\left(\frac{x}{2}\right)=\frac{1}{2}\cos\left(\frac{x}{2}\right)-\frac{1}{2}\cos\left(\frac{3x}{2}\right)
 $$ and we have done.
A: An easy way to do that is:
sinx = sqrt((1+cosx)(1-cosx))
so LHS becomes:
cosx + sqrt((1+cosx)(1-sinx))sqrt((1-cosx)/1+cosx))
cosx + 1 - cosx = 1 = RHS
A: $ t = \tan(../2) $
$$ \sin ( ..) = \frac{ 2 t }{1+t^2}  ; \,  \cos ( ..) = \frac{ 1-t^2 }
{1+t^2} $$
Plug in
$$\cos x+\sin x\tan \frac{x}{2} \rightarrow 1 $$
