Question
Given that $$a, b, c$$ are the sides of a triangle $$ABC$$ which is right angled at $$C,$$ then the minimum value of $${\left( {\frac{c}{a} + \frac{c}{b}} \right)^2}$$ is
A.
0
B.
4
C.
6
D.
8
Answer :
8
Solution :
$$\eqalign{ & a = c\sin \theta ,b = c\cos \theta \cr & \Rightarrow {\left( {\frac{c}{a} + \frac{c}{b}} \right)^2} \cr & = {\left( {\frac{1}{{\sin \theta }} + \frac{1}{{\cos \theta }}} \right)^2} = \frac{{4\left( {1 + \sin 2\theta } \right)}}{{{{\sin }^2}2\theta }} \cr & = 4\left( {\frac{1}{{{{\sin }^2} 2\theta }} + \frac{1}{{\sin 2\theta }}} \right),{\text{where }}\,0 < \theta < \frac{\pi }{2} \cr & \Rightarrow {\left( {\frac{c}{a} + \frac{c}{b}} \right)^2}_{\min } = 8,{\text{when }}\,2\theta = {90^ \circ }. \cr} $$
$$\eqalign{ & a = c\sin \theta ,b = c\cos \theta \cr & \Rightarrow {\left( {\frac{c}{a} + \frac{c}{b}} \right)^2} \cr & = {\left( {\frac{1}{{\sin \theta }} + \frac{1}{{\cos \theta }}} \right)^2} = \frac{{4\left( {1 + \sin 2\theta } \right)}}{{{{\sin }^2}2\theta }} \cr & = 4\left( {\frac{1}{{{{\sin }^2} 2\theta }} + \frac{1}{{\sin 2\theta }}} \right),{\text{where }}\,0 < \theta < \frac{\pi }{2} \cr & \Rightarrow {\left( {\frac{c}{a} + \frac{c}{b}} \right)^2}_{\min } = 8,{\text{when }}\,2\theta = {90^ \circ }. \cr} $$