A) 0
B) 1
C) 2
D) \[\tan A\]
Correct Answer: A
Solution :
Here \[\tan 2A=\frac{2\sqrt{\frac{{{k}^{2}}}{4}+{{\tan }^{2}}A}}{1-{{\tan }^{2}}A}\] \[\Rightarrow \frac{2\tan A}{1-{{\tan }^{2}}A}=\frac{2\sqrt{\frac{{{k}^{2}}}{4}+{{\tan }^{2}}A}}{1-{{\tan }^{2}}A}\] \[\Rightarrow \frac{{{k}^{2}}}{4}+{{\tan }^{2}}A={{\tan }^{2}}A\Rightarrow k=0\] .You need to login to perform this action.
You will be redirected in
3 sec