A) 1
B) ? 1
C) 0
D) None of these
Correct Answer: A
Solution :
Let \[{{y}_{1}}={{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}}\] and \[{{y}_{2}}={{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}\] Differentiating w.r.t. x of\[{{y}_{1}}\]and \[{{y}_{2}}\], we get \[\frac{d{{y}_{1}}}{dx}=\frac{d}{dx}\left[ {{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}} \right]\] Putting \[x=\tan \theta \], \[\therefore {{y}_{1}}={{\tan }^{-1}}\tan 2\theta =2\theta =2{{\tan }^{-1}}x\] and \[{{y}_{2}}={{\sin }^{-1}}\sin 2\theta =2{{\tan }^{-1}}x\] Again \[\frac{d{{y}_{1}}}{dx}=\frac{d}{dx}[2{{\tan }^{-1}}x]=\frac{2}{1+{{x}^{2}}}\] .....(i) and \[\frac{d{{y}_{2}}}{dx}=\frac{d}{dx}[2{{\tan }^{-1}}x]=\frac{2}{1+{{x}^{2}}}\] .....(ii) Hence \[\frac{d{{y}_{1}}}{d{{y}_{2}}}=1\].
You need to login to perform this action.
You will be redirected in
3 sec
