Question

If $x=sec\Phi -\tan \Phi ,y=\cos ec\Phi +cot\Phi$, 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

$x=\frac{y-1}{y+1}$

Explanation for the correct option.

Step 1:Simplify $x$

$\begin{array}{rcl}x& =& sec\varphi -\tan \varphi \\ & =& \frac{1}{\cos \varphi }-\frac{\sin \varphi }{\cos \varphi }\\ & =& \frac{1-\sin \varphi }{\cos \varphi }\end{array}$

Step 2:Simplify $y$

$\begin{array}{rcl}y& =& \cos ec\varphi +cot\varphi \\ & =& \frac{1}{\sin \varphi }+\frac{\cos \varphi }{\sin \varphi }\\ & =& \frac{1+\cos \varphi }{\sin \varphi }\end{array}$

Step 3: Find the value of $xy+x-y+1$

$\begin{array}{rcl}xy+x-y+1& =& \left(\frac{1-\sin \varphi }{\cos \varphi }\times \frac{1+\cos \varphi }{\sin \varphi }\right)+\frac{1-\sin \varphi }{\cos \varphi }-\frac{1+\cos \varphi }{\sin \varphi }+1\\ & =& \left(\frac{1+\cos \varphi -\sin \varphi -\cos \varphi \sin \varphi }{\cos \varphi \sin \varphi }\right)+\frac{1}{\cos \varphi }-\frac{\sin \varphi }{\cos \varphi }-\frac{1}{\sin \varphi }-\frac{\cos \varphi }{\sin \varphi }+1\\ & =& \frac{1}{\cos \varphi \sin \varphi }+\frac{1}{\sin \varphi }-\frac{1}{\cos \varphi }-1+\frac{1}{\cos \varphi }-\frac{\sin \varphi }{\cos \varphi }-\frac{1}{\sin \varphi }+\frac{\cos \varphi }{\sin \varphi }+1\\ & =& \frac{1}{\cos \varphi \sin \varphi }-\frac{\sin \varphi }{\cos \varphi }+\frac{\cos \varphi }{\sin \varphi }\\ & =& \frac{1-\left({\sin }^2\varphi +\cos {}^2\varphi \right)}{\cos \varphi \sin \varphi }\\ & =& 0\end{array}$

So, $\begin{array}{rcl}xy+x-y+1& =& 0\\ \Rightarrow x\left(y+1\right)& =& y-1\\ \Rightarrow x& =& \frac{y-1}{y+1}\end{array}$

Hence, option B is correct.