Question

If $ax+b\left(\sec \right({\tan }^{-1}x\left)\right)=c$ and $ay+b\left(\sec \right({\tan }^{-1}y\left)\right)=c$, then the value of $\frac{x+y}{1-xy}$ is,

• A. $\frac{2ab}{a^2-c^2}$
• B. $\frac{2ac}{a^2-c^2}$
• C. $\frac{c^2-b^2}{a^2+b^2}$
• D. none of these

Answer 1

The correct option is B $\frac{2ac}{a^2-c^2}$

$ax+b\left(sec\left(ta{n}^{-1}x\right)\right)=c$

$\Rightarrow ax+b\left(sec\left(se{c}^{-1}\sqrt{1+x^2}\right)\right)=c$

$\Rightarrow ax+b\sqrt{1+x^2}=c\Rightarrow b\sqrt{1+x^2}=c-ax$

$\Rightarrow b^2(1+x^2)=c^2+a^2{x}^2-2acx$

$\Rightarrow (b^2-a^2)x^2+(b^2-c^2)+2acx=0$

$\Rightarrow x=\frac{-2ac\pm \sqrt{4a^2{c}^2-4(b^2-a^2)(b^2-c^2)}}{2(b^2-a^2)}$

Similarly

$ay+bsec\left(ta{n}^{-1}y\right)=c$

$\Rightarrow y=\frac{-2ac\pm \sqrt{4a^2{c}^2-4(b^2-a^2)(b^2-c^2)}}{2(b^2-a^2)}$

Gives
$xy=\frac{-2ac+\sqrt{4a^2{c}^2-4(b^2-a^2)(b^2-c^2)}}{2(b^2-a^2)}\times \frac{-2ac-\sqrt{4a^2{c}^2-4(b^2-a^2)(b^2-c^2)}}{2(b^2-a^2)}$

$=\frac{4a^2{c}^2-4a^2{c}^2-4(b^2-a^2)(b^2-c^2)}{4{(b^2-a^2)}^2}=\frac{c^2-b^2}{a^2-b^2}$

And $x+y=\frac{2ac}{(a^2-b^2)}$

Therefore,

$\frac{x+y}{1-xy}=\frac{2ac}{a^2-c^2}$