Question

If ${\sec }^{-1}\left(\frac{1+x}{1-y}\right)=a$, then $\frac{dy}{dx}$ is equal to

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

Answer 1

The correct option is A

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

Find $\frac{dy}{dx}$ by simplifying the given function:

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

After cross-multiplication, we get

$\begin{array}{rcl}\therefore \left(1-y\right)\sec \left(a\right)& =& 1+x\\ & \Rightarrow & \sec \left(a\right)-y\sec \left(a\right)=1+x\\ & \Rightarrow & y\sec \left(a\right)=\sec \left(a\right)-1-x\end{array}$

Now, differentiate both the sides with respect to $x$,

$\begin{array}{rcl}\therefore \frac{dy}{dx}\sec \left(a\right)+y\frac{\mathrm{d}}{dx}\left(\sec \left(a\right)\right)& =& \frac{\mathrm{d}}{dx}\left(\sec \left(a\right)\right)-0-1\\ & \Rightarrow & \frac{dy}{dx}\sec \left(a\right)+y\left(0\right)=0-1\\ & \Rightarrow & \frac{dy}{dx}=-\frac{1}{\sec \left(a\right)}\end{array}$

Since $\sec a=\frac{1+x}{1-y}$, then

$\begin{array}{rcl}\frac{dy}{dx}& =& -\frac{1}{{\displaystyle \frac{1+x}{1-y}}}\\ & =& -\frac{1-y}{1+x}\\ & =& \frac{y-1}{x+1}\end{array}$

Hence, the correct option is (A).