Question

If $x=\sec \varphi -\tan \varphi$ 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 $xy+x-y+1=0$
D $y=\frac{1+x}{1-x}$

$x=\sec \varphi -\tan \varphi \Rightarrow x=\frac{1-\sin \varphi }{\cos \varphi }=\frac{1-\cos (\frac{\pi }{2}-\varphi )}{\sin (\frac{\pi }{2}-\varphi )}\Rightarrow x=\frac{2{\sin }^2(\frac{\pi }{4}-\frac{\varphi }{2})}{2\sin (\frac{\pi }{4}-\frac{\varphi }{2})\cos (\frac{\pi }{4}-\frac{\varphi }{2})}$
$\Rightarrow x=\tan (\frac{\pi }{4}-\frac{\varphi }{2})=\frac{1-\tan \frac{\varphi }{2}}{1+\tan \frac{\varphi }{2}}$ ...(1)
$y=\csc \varphi +\cot \varphi =\frac{1+\cos \varphi }{\sin \varphi }\Rightarrow y=\frac{2{\cos }^2\frac{\varphi }{2}}{2\sin \frac{\varphi }{2}\cos \frac{\varphi }{2}}$
$\Rightarrow y=\cot \frac{\varphi }{2}$ ...(2)
From (1) & (2)
$x=\frac{y-1}{y+1}$
$\Rightarrow xy+x-y+1=0$
$\Rightarrow y=\frac{x+1}{x-1}$
Ans: B,C,D