Question

If $x=\sec \varphi -\tan \varphi$ & $y=cosec\varphi +\cot \varphi$, then

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

Answer 1

The correct option is A $x=\frac{y-1}{y+1}$

Given:$x=\sec \varphi -\tan \varphi$ and $y=cosec\varphi +\cot \varphi$

Now, $xy=$ ( $sec$ { $\varphi$ } $-tan$ { $\varphi$ } )(

$cosec$ { $\varphi$ } +$cot$ { $\varphi$ } )

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