Question

If $ax+b\left(\sec \left({\tan }^{-1}x\right)\right)=canday+b\left(\sec \left({\tan }^{-1}y\right)\right)=c,then\frac{x+y}{1-xy}=$

• A. $\frac{ac}{a^2+c^2}$
• B. $\frac{2ac}{a-c}$
• C. $\frac{2ac}{a^2-c^2}$
• D. $\frac{a+c}{1-ac}$

Answer 1

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

Comparing both the equations,

$ax+b\sec {\tan }^{-1}x=c$ and $ay+b\sec {\tan }^{-1}y=c$ we have

$\Rightarrow ax+b\sec {\tan }^{-1}x=ay+b\sec {\tan }^{-1}y$

$\Rightarrow ax+b\sec {\tan }^{-1}x-ay-b\sec {\tan }^{-1}y=0$

$\Rightarrow ax-ay+b\sec {\tan }^{-1}x-b\sec {\tan }^{-1}y=0$

$\Rightarrow ax-ay=0$ or $b(\sec {\tan }^{-1}x-\sec {\tan }^{-1}y)=0$

$\Rightarrow ax=ay$ or $b\ne 0,\sec {\tan }^{-1}x-\sec {\tan }^{-1}y=0$

$\Rightarrow x=y$ or $\sec {\tan }^{-1}x-\sec {\tan }^{-1}y=0$

Let $x=\tan \theta$

Consider $ax+b\sec {\tan }^{-1}x=c$

$\Rightarrow a\tan \theta +b\sec {\tan }^{-1}\tan \theta =c$

$\Rightarrow a\tan \theta +b\sec \theta =c$ ......$\left(1\right)$

$\Rightarrow a\frac{\sin \theta }{\cos \theta }+b\frac{1}{\cos \theta }=c$

$\Rightarrow a\sin \theta +b=c\cos \theta$

or $c\cos \theta -a\sin \theta =b$ ......$\left(2\right)$

Let $a^2+c^2=r^2$ for some $\alpha$

where $a=r\cos \alpha ,a=r\sin \alpha$

then $\tan \alpha =\frac{a}{c}$

Thus,$\cos \alpha \cos \theta -\sin \alpha \sin \theta =\frac{b}{r}$

$\Rightarrow \cos (\alpha +\theta )=\frac{b}{r}$

$\Rightarrow \alpha +\theta =\pm {\cos }^{-1}\frac{b}{r}$

Let $\alpha +\theta$ be the positive solution, and $\alpha +\theta$ the negative solution, where $y=\tan \varphi$

$\Rightarrow \alpha +\varphi =-(\alpha +\theta )$

$\Rightarrow -2\alpha =\theta +\varphi$

Taking tangents, both sides,we get

$\tan (-2\alpha )=\tan (\theta +\varphi )$

$\frac{-2\tan \alpha }{1-{\tan }^2\alpha }=\frac{\tan \theta +\tan \varphi }{1-\tan \theta \tan \varphi }$

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

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

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