Problems based on triangles and trigonometry. In an acute triangle $x,y,z$ are the given angles where $\cos x=\tan y$, $\cos y = \tan z$ and $\cos z = \tan x$. Find the sum of sines in the triangle.
Could be done by substituting values of $\sin$ function but in vain. Can anyone please help me?
 A: Hint: 
Let $a=\tan^2 x, b =\tan^2y, c=\tan^2 z$. By squaring and using $\cos^2x =\frac{1}{1+\tan^2x}$, we have $$ b=\frac{1}{1+a} \\ c=\frac{1}{1+b} \\ a=\frac{1}{1+c}$$ By substituting $c$ and $b$ you get a quadratic in $a$ : $$a^2+a-1=0 \implies a=\frac{\sqrt 5-1}{2}$$ and from here you get $$\sin x =\sqrt{\frac{\sqrt 5-1}{\sqrt 5+1}}$$ by some simple trig identities. You can find $\sin y$ and $\sin z$ in a similar way.
A: The question states that it's given that
in an acute triangle 
$x,y,z$ are the given angles where
\begin{align} 
\cos x&=\tan y
\tag{1}\label{1}
,\\
\cos y&=\tan z
\tag{2}\label{2}
,\\
\cos z&=\tan x
\tag{3}\label{3}
. 
\end{align} 
There are many ways to prove that 
there is no valid triangle with such properties.
For one is the incomplete answer
which implies that 
it follows from \eqref{1}-\eqref{3}
that 
\begin{align} 
\tan x&=\tan y=\tan z=
\cos x=\cos y=\cos z=
=\sqrt{\tfrac12\,(\sqrt 5-1)}
,
\end{align}
which is absurd.
Another way is:
rewriting \eqref{1}-\eqref{3} as
\begin{align} 
\cos x\cos y&=\sin y
\tag{4}\label{4}
,\\
\cos y\cos z&=\sin z
\tag{5}\label{5}
,\\
\cos z\cos x&=\sin x
\tag{6}\label{6}
, 
\end{align}
so
\begin{align}
\sin x+\sin y+\sin z
&= 
\cos x\cos y+\cos y\cos z+\cos z\cos x
\tag{7}\label{7}
.
\end{align}
Using known identities
\begin{align} 
\sin x+\sin y+\sin z&=u
\tag{8}\label{8}
\end{align}
and
\begin{align} 
\cos x\cos y+\cos y\cos z+\cos z\cos x
&=\frac{u^2+v^2}4-1
\tag{9}\label{9}
,
\end{align}
where $u=\rho/R$, $v=r/R$
and $\rho,r,R$ are the semiperimeter,
inradius and circumradius of given triangle (if such exists).
From equations \eqref{7}-\eqref{9},
\begin{align}
\frac{u^2+v^2}4-1=v
,\\
u&=2+\sqrt{8-v^2}
,
\end{align} 
and this expression for
\begin{align}
u>\tfrac{3\sqrt3}2
&=\max_{v\in[0,1/2]}u(v)
,
\end{align}
that is, there is no a pair $(u,v)$
that simultaneously agree with \eqref{7}-\eqref{9}
and represent a valid triangle.
