question_answer

Given, that a, b, c are the sides of a triangle ABC which is right angles 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

Correct Answer: D

Solution :

[d] \[a=c\sin \theta ,b=c\,\,cos\theta \] \[\Rightarrow {{\left( \frac{c}{a}+\frac{c}{b} \right)}^{2}}\] \[={{\left( \frac{1}{\sin \theta }+\frac{1}{\cos \theta } \right)}^{2}}=\frac{4(1+sin2\theta )}{{{\sin }^{2}}2\theta }\] \[=4\left( \frac{1}{{{\sin }^{2}}2\theta }+\frac{1}{\sin 2\theta } \right),\] Where \[0<\theta <\frac{\pi }{2}\] \[\Rightarrow \,\,\,\,\,{{\left( \frac{c}{a}+\frac{c}{b} \right)}^{2}}_{\min }=8,\] when \[2\theta =90{}^\circ .\]