A) \[\frac{2a}{1+{{a}^{2}}}\]
B) \[\frac{1-{{a}^{2}}}{1+{{a}^{2}}}\]
C) \[\frac{2a}{1-{{a}^{2}}}\]
D) None of these
Correct Answer: C
Solution :
\[\tan \left[ \frac{1}{2}{{\sin }^{-1}}\left( \frac{2a}{1+{{a}^{2}}} \right)+\frac{1}{2}{{\cos }^{-1}}\left( \frac{1-{{a}^{2}}}{1+{{a}^{2}}} \right) \right]\] \[=\tan \left[ \frac{1}{2}{{\sin }^{-1}}\left( \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)+\frac{1}{2}{{\cos }^{-1}}\left( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right) \right]\] (Let\[a=\tan \theta \]) \[=\tan \left[ \frac{1}{2}{{\sin }^{-1}}(\sin 2\theta )+\frac{1}{2}{{\cos }^{-1}}(\cos 2\theta ) \right]\] \[=\tan (2\theta )=\tan 2\theta =\frac{2\tan \theta }{1-{{\tan }^{2}}\theta }=\frac{2a}{1-{{a}^{2}}}\] Trick: Put\[a=0\], then tan (0+0) = 0; which is given by (a) and (c). Again put\[a=1\], then \[\tan \left( \frac{\pi }{4}+\frac{\pi }{4} \right)=\infty \], which is given by (c).
You need to login to perform this action.
You will be redirected in
3 sec
