Question

lf $x=\sec \varphi -\tan \varphi$ and $y=\text{cosec}\varphi +\cot \varphi$, then

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

Answer 1

Answer: (B), (C), (D)

The correct options are
B $x=\frac{y-1}{y+1}$
C $y=\frac{1+x}{1-x}$
D $xy+x-y+1=0$

$x=\sec \varphi -\tan \varphi =\frac{1}{\cos \varphi }-\frac{\sin \varphi }{\cos \varphi }=\frac{1-\sin \varphi }{\cos \varphi }y=\text{cosec}\varphi +\cot \varphi =\frac{1}{\sin \varphi }+\frac{\cos \varphi }{\sin \varphi }=\frac{1+\cos \varphi }{\sin \varphi }xy+x-y+1=\frac{1-\sin \varphi }{\cos \varphi }.\frac{1+\cos \varphi }{\sin \varphi }+\frac{1-\sin \varphi }{\cos \varphi }-\frac{1+\cos \varphi }{\sin \varphi }+1=\frac{1-\sin \varphi +\cos \varphi -\sin \varphi \cos \varphi +\sin \varphi -{\sin }^2\varphi -\cos \varphi -{\cos }^2\varphi +\sin \varphi \cos \varphi }{\sin \varphi \cos \varphi }=0$
$xy+x-y+1=0\Rightarrow x(y+1)=y-1\Rightarrow x=\frac{y-1}{y+1}\Rightarrow y=\frac{1+x}{1-x}$