Question

If $x=\sec \varphi -\tan \varphi$ and $y=cosec\varphi$ then:

• A. $x=\frac{y+1}{y-1}$
• B. $x=\frac{y-2}{x-2}$
• C. $xy+x+y+1=0$
• D. $y{x}^2+x^2-y+1=0$

Answer 1

The correct option is D $y{x}^2+x^2-y+1=0$

$x=sec\varphi -tan\varphi$ and $se{c}^2\varphi -ta{n}^2\varphi =1$

$\frac{1}{x}=\frac{1}{(sec\varphi -tan\varphi )}$ now multiply and divide by $sec\varphi +tan\varphi$

$\frac{1}{x}=\frac{sec\varphi +tan\varphi )}{(se{c}^2\varphi -ta{n}^2\varphi )}=sec\varphi +tan\varphi$

$x+\frac{1}{x}=2sec\varphi$ and $x-\frac{1}{x}=-2tan\varphi$

$y=cosec\varphi =sec\varphi /tan\varphi =\frac{(x+\frac{1}{x})}{-(x-\frac{1}{x})}=\frac{1+x^2}{1-x^2}$

$y-y{x}^2=1+x^2=>x^2+x^{2}y-y+1=0$