A) 2
B) -2
C) 1
D) 3
Correct Answer: B
Solution :
Given, \[\cos \theta =\frac{2}{3}=\frac{Base}{Hypotenuse}\] |
Let \[AB=2k\] and \[BC=3k\] |
Apply Pythagoras theorem, |
\[A{{B}^{2}}+A{{C}^{2}}=B{{C}^{2}}\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{\left( 2k \right)}^{2}}=A{{C}^{2}}+{{\left( 3k \right)}^{2}}\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,A{{C}^{2}}=9{{k}^{2}}-4{{k}^{2}}\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,AC=\sqrt{5}k\] |
\[\sec \theta =\frac{Hypotemuse}{Base}=\frac{3}{2},\tan \theta =\frac{p}{b}=\frac{\sqrt{5}}{2}\] |
\[2{{\sec }^{2}}\theta +2{{\tan }^{2}}\theta -9\] |
\[=2\times {{\left( \frac{3}{2} \right)}^{2}}+2\times {{\left( \frac{\sqrt{5}}{2} \right)}^{2}}-9\] |
\[=2\times \frac{9}{4}+\frac{2\times 5}{4}-9=\frac{18+10-36}{4}\] |
\[=\frac{28-36}{4}=-\frac{8}{4}=-2\] |
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