Question

If $x=sec\theta +\tan \theta ,y=\csc \theta -\cot \theta$ then $y=$

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

Answer 1

The correct option is D $\frac{x-1}{x+1}$

$x=sec\theta +\tan \theta ........\left(i\right)$
$y=\csc \theta -\cot \theta ...........\left(ii\right)$
$multiply\left(i\right)with\left(ii\right)$
$xy=(\sec \theta +\tan \theta )(\csc \theta -\cot \theta )$
$=\sec \theta \csc \theta -\sec \theta \cot \theta +\tan \theta \csc \theta -\tan \theta \cot \theta$
$=\sec \theta \csc \theta -\sec \theta \cot \theta +\tan \theta \csc \theta -1[\cot \theta =\frac{1}{\tan \theta }]$
$=\frac{1}{\cos \theta \sin \theta }-\frac{1}{\cos \theta }\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{\cos \theta }\frac{1}{\sin \theta }-1$
$=\frac{{\sin }^2\theta +{\cos }^2\theta }{\cos \theta \sin \theta }-\csc \theta +\sec \theta -1$
$=\tan \theta +\cot \theta -\csc \theta +\sec \theta -1$
$=\tan \theta +\sec \theta -(\csc \theta -\cot \theta )-1$
$xy=x-y-1$
$\Rightarrow xy+y=x-1$
$\Rightarrow y(x+1)=x-1$
$y=\frac{x-1}{x+1}$