If tan-1a+xa+tan-1a-xa=π6 then x2 is equal to
23a
3a
23a2
None of these
Explanation for the correct option
Given that tan-1a+xa+tan-1a-xa=π6
⇒tan-1a+xa+a-xa1-a+xa×a-xa=π6;∵tan-1x+tan-1y=tan-1x+y1-xy⇒tan-12aaa2-a2-x2a2=π6⇒tan-12a2x2=π6⇒2a2x2=tanπ6⇒2a2x2=13⇒x2=23a2
Hence the correct option is option(C) i.e. 23a2