Q.
In a $\triangle$ ABC w ith fixed base BC, the vertex A moves
such that cos B + cos C = 4 $ sin^2 \frac{A}{2}$. If a, b and c denote
th e lengths of th e sides of th e triangle opposite to the
angles A, B and C respectively, then
IIT JEEIIT JEE 2009Trigonometric Functions
Solution:
Given, cos B + cos C = 4 $ sin^2 \frac{A}{2} $
$\Rightarrow 2 \, cos \, \bigg( \frac{ B + C }{2} \bigg) \, cos \, \bigg( \frac{ B - C }{2} \bigg) = 4 \, sin^2 \frac{A}{2} $
$\Rightarrow 2 \, sin \, \frac{A}{2} \bigg [ cos \bigg( \frac{ B - C }{2} \bigg) - 2 \, sin \frac{A}{2} \bigg ] = 0 $
$\Rightarrow cos \bigg( \frac{ B - C }{2} \bigg) - 2 \, cos \bigg( \frac{ B + C }{2} \bigg) = 0 $
as sin $ \frac{A}{2} \ne 0 $
$\Rightarrow - cos \frac{ B }{2} \, cos \, \frac{C }{2} + 3 sin \, \frac{B}{2} \, sin \, \frac{C }{2} = 0 $
$\Rightarrow tan \frac{B }{2} tan \, \frac{C}{2} = \frac{1}{3} $
$\Rightarrow \sqrt{ \frac{ ( s - a) \, (s - c) }{ s \, (s - b) } . \frac{( s - b ) \, (s - a)}{ s \, (s - c)}} \frac{1}{3} $
$\Rightarrow \frac{s - a}{s} \frac{1}{3} $
$\Rightarrow 2s = 3a $
$\Rightarrow b + c = 2a $.
$\therefore $ Locus of A is an ellipse
