If \sqrt {a+b} \tan \frac {\theta}{2}=\sqrt {a-b} \tan \frac {\phi}{2}
then prove that(b-a\ sec \theta)(b+a\cos \phi)\leq 0
-
UP 0 DOWN 0 0 1
1 Answers
Lokesh Verma
·2009-11-09 06:25:26
I can see something funny here:
take b/a = x
prove that there is one root of x between sec θ and - cos φ