Trigonometry! #2

Well I was dumb and clicked the wrong button, but here's my alternative solution using pure Geometry since I hate trig. Lol

$\sec(x)=\dfrac{c}{b}, ~~~~~~ \tan(x)=\dfrac{a}{b}$

$\sec(x)-\tan(x)=\frac{c}{b}-\frac{a}{b}=\frac{c-a}{b}=\frac{1}{p}$

Using some logic and the ever useful powers of laziness, we let $a=c-1$ and $b=p$

$\frac{c-a}{b}\Rightarrow \frac{c-(c-1)}{p}=\frac{1}{p}$

Using Pythagorean theorem

$c^2=a^2+b^2\Rightarrow (c-1)^2+p^2$

$c^2=c^2-2c+1+p^2$

$2c-1=p^2$

$c=\frac{p^2+1}{2}$

Since earlier we let $a=c-1$ we substitute for c to get

$a=(\frac{p^2+1}{2})-1=\frac{p^2-1}{2}$

Now, $\csc{x}=\frac{c}{a}$ , substituting our values for c and a in terms of p we get

$\csc{x}=\frac{c}{a}\Rightarrow \cfrac{\frac{p^2+1}{2}}{\frac{p^2-1}{2}}= \dfrac{p^2+1}{p^2-1}$

I don't see any significance in taking $\theta = x$ though. :\

Good solution ...I did same but the I selected the option of sin (theta) instead of cosec(theta) by mistake......:-)

I pressed the answer you have solved but it showed that I was wrong . Do something about it !!!!

A very elegant solution!!