Question

If $x=\tan \theta +\cot \theta ,y=\cos \theta -\sin \theta$, then which of the following is true?

• 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

Answer: (B)

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

Given, $x=\tan \theta +\cot \theta$

$\Rightarrow x=\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }$

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

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

$\Rightarrow \sin \theta \cos \theta =\frac{1}{x}$ ......(i)

Also given, $y=\cos \theta -\sin \theta$

On squaring both sides, we have

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

On substituting value of (i), we get

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

So, option B is correct.