Inverse Trigonometric Functions Question 70
Question: If $ ax+b(sec(ta{n^{-1}}x))=c $ and $ ay+b $
$ (sec.(ta{n^{-1}}y))=c, $ then $ \frac{x+y}{1-xy}= $
Options:
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} $
Show Answer
Answer:
Correct Answer: C
Let $ {{\tan }^{-1}}x=\alpha $ and $ {{\tan }^{-1}}y=\beta \Rightarrow \tan \alpha =x,\tan \beta =y $ .
The given system of equations is $ a\tan \alpha +b\sec \alpha =c $ and $ a\tan \beta +b\sec \beta =c $
$ \therefore \alpha and\beta $ are the roots of $ a\tan \theta +b\sec \theta =c $
$ \Rightarrow {{(bsec\theta )}^{2}}={{(c-atan\theta )}^{2}} $
$ \Rightarrow (a^{2}-b^{2})tan^{2}\theta -2ac\tan \theta +c^{2}-b^{2}=0 $
$ \Rightarrow \tan \alpha +tan\beta =\frac{2ac}{a^{2}-b^{2}} $ And $ \tan \alpha \tan \beta =\frac{c^{2}-b^{2}}{a^{2}-b^{2}} $
$ \Rightarrow x+y=\frac{2ac}{a^{2}-b^{2}} $ and $ xy=\frac{c^{2}-b^{2}}{a^{2}-b^{2}} $
$ \Rightarrow 1-xy=\frac{a^{2}-c^{2}}{a^{2}-b^{2}}\Rightarrow \frac{x+y}{1-xy}=\frac{2ac}{a^{2}-c^{2}} $
BETA
