Question

Eliminate $z$ from
$p\csc hz+q\sec hz+r=0$
$p_{\prime }\csc hz+q_{\prime }\sec hz+r^{\prime }=0$

Answer 1

Solving by the Rule of Cross Multiplication, we get
$p\csc hz+q\sec hz+r=0$
$p^{\prime }\csc hz+q^{\prime }\sec hz+r^{\prime }=0$
$\sec hz$ $1$ $\csc hz$
$q$ $r$ $p$ $q$
$q^{\prime }$ $r^{\prime }$ $p^{\prime }$ $q^{\prime }$
$\frac{\csc hz}{q{r}^{\prime }-q^{\prime }r}=\frac{\sec hz}{r{p}^{\prime }-r^{\prime }p}=\frac{1}{p{q}^{\prime }-q{p}^{\prime }}$
$\Rightarrow \frac{\csc hz}{q{r}^{\prime }-q^{\prime }r}=\frac{1}{p{q}^{\prime }-q{p}^{\prime }}$ and $\frac{\sec hz}{r{p}^{\prime }-r^{\prime }p}=\frac{1}{p{q}^{\prime }-q{p}^{\prime }}$
$\Rightarrow \csc hz=\frac{q{r}^{\prime }-q^{\prime }r}{p{q}^{\prime }-q{p}^{\prime }}$ and $\sec hz=\frac{r{p}^{\prime }-r^{\prime }p}{p{q}^{\prime }-q{p}^{\prime }}$
$\sin hz=\frac{p{q}^{\prime }-q{p}^{\prime }}{q{r}^{\prime }-q^{\prime }r}$
and $\cos hz=\frac{p{q}^{\prime }-q{p}^{\prime }}{r{p}^{\prime }-r^{\prime }p}$
Squaring and subtracting we get
${\cos }^{2}hz-{\sin }^{2}hz={\left(\frac{p{q}^{\prime }-q{p}^{\prime }}{r{p}^{\prime }-r^{\prime }p}\right)}^2-{\left(\frac{p{q}^{\prime }-q{p}^{\prime }}{q{r}^{\prime }-q^{\prime }r}\right)}^2$
$\Rightarrow {\left(\frac{p{q}^{\prime }-q{p}^{\prime }}{r{p}^{\prime }-r^{\prime }p}\right)}^2-{\left(\frac{p{q}^{\prime }-q{p}^{\prime }}{q{r}^{\prime }-q^{\prime }r}\right)}^2=1$