Question

If $x=\tan \theta +\cot \theta ,y=\cos \theta -\sin \theta$, then

• A. $x=y$
• B. $\frac{1-y^2}{2}=\frac{1}{x}$
• C. $\frac{y^2-1}{2}=\frac{1}{x}$
• D. $\frac{1+y^2}{2}=\frac{1}{x}$

Answer 1

The correct option is B $\frac{1-y^2}{2}=\frac{1}{x}$

$x=\tan \theta +\cot \theta =\frac{\sin \theta }{cos\theta }+\frac{\cos \theta }{\sin \theta }$

$=\frac{{\cos }^2\theta +{\sin }^2\theta }{\cos \theta \sin \theta }=\frac{1}{\cos \theta \sin \theta }$

$y=\cos \theta -\sin \theta$

$\Rightarrow \frac{1-y^2}{2}=\frac{1-(\cos \theta -\sin \theta {)}^2}{2}=\frac{1-({\cos }^2\theta +{\sin }^2\theta -2\sin \theta \cos \theta )}{2}$

$=\frac{1-(1-2\sin \theta \cos \theta )}{2}=\frac{1-1+2\sin \theta \cos \theta }{2}$

$=\frac{2\sin \theta \cos \theta }{2}=\sin \theta \cos \theta$

and $\frac{1}{x}=\sin \theta \cos \theta$

$\therefore \frac{1-y^2}{2}=\frac{1}{x}$