question_answer

If \[\cos \theta =\frac{2}{3}\], then the value of \[2{{\sec }^{2}}\theta +2{{\tan }^{2}}\theta -9\] is

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,

\[A{{B}^{2}}+A{{C}^{2}}=B{{C}^{2}}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{\left( 2k \right)}^{2}}=A{{C}^{2}}+{{\left( 3k \right)}^{2}}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,A{{C}^{2}}=9{{k}^{2}}-4{{k}^{2}}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,AC=\sqrt{5}k\]

\[\sec \theta =\frac{Hypotemuse}{Base}=\frac{3}{2},\tan \theta =\frac{p}{b}=\frac{\sqrt{5}}{2}\]

\[2{{\sec }^{2}}\theta +2{{\tan }^{2}}\theta -9\]

\[=2\times {{\left( \frac{3}{2} \right)}^{2}}+2\times {{\left( \frac{\sqrt{5}}{2} \right)}^{2}}-9\]

\[=2\times \frac{9}{4}+\frac{2\times 5}{4}-9=\frac{18+10-36}{4}\]

\[=\frac{28-36}{4}=-\frac{8}{4}=-2\]