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. None of these

Answer 1

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

To solve the problem in minimum tru take $b=\frac{\pi }{4}$

look into options

$x=\sqrt{2}-1y=\sqrt{2}+1$

$\left(A\right)\frac{y+1}{y-1}=\frac{\sqrt{2}+2}{\sqrt{2}}=\sqrt{2}+1\ne x$

$\left(B\right)\frac{y-1}{y+1}=\frac{\sqrt{2}}{\sqrt{2}+2}=\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{2-1}=x$

so $B$ is correct

$\left(C\right)\frac{1-x}{1+x}=\frac{1-(\sqrt{2}-1)}{1+\sqrt{2}-1}=\frac{2-\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}-1}{-y}$