\[\Delta ABC\]is right angled triangle, where \[\angle ABC=90{}^\circ .\]If \[AC=2\sqrt{5}\]and \[AB-BC=2,\] then the value of \[{{\cos }^{2}}A-{{\cos }^{2}}C\]is [SSC (CGL) Mains 2012] |
A) \[\frac{1}{\sqrt{5}}\]
B) \[\sqrt{5}\]
C) \[\frac{1}{2}\]
D) \[\frac{3}{5}\]
Correct Answer: D
Solution :
Let \[BC=x,\]then \[AB=x+2\] |
In \[\Delta ABC,\] |
\[{{(x+2)}^{2}}+{{x}^{2}}={{(2\sqrt{5})}^{2}}\] |
\[\Rightarrow \]\[{{x}^{2}}+4+4x+{{x}^{2}}=20\] |
\[\Rightarrow \]\[2{{x}^{2}}+4x-16=0\] |
\[\Rightarrow \]\[{{x}^{2}}+2x-8=0\] |
\[\Rightarrow \]\[(x-2)(x+4)=0\] |
\[\Rightarrow \] \[x=2\] |
So, \[AB=4,\]\[BC=2\] |
\[\therefore \]\[{{\cos }^{2}}A-{{\cos }^{2}}C={{\left( \frac{4}{2\sqrt{5}} \right)}^{2}}-{{\left( \frac{2}{2\sqrt{5}} \right)}^{2}}=\frac{3}{5}\] |
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