Question

If $\cos \theta =\dfrac{2}{3}$, then the value of $2{{\sec }^{2}}\theta +2{{\tan }^{2}}\theta -7$ is equal to
[a] 1
[b] 0
[c] 3
[d] 4

Answer 1

Hint: Use the fact that if cosx =a then $\tan x=\pm \dfrac{\sqrt{1-{{a}^{2}}}}{a}$. Use the formula that ${{\sec }^{2}}x=1+{{\tan }^{2}}x$. Hence find the value of ${{\sec }^{2}}x$ and ${{\tan }^{2}}x$ and hence find the value of the given expression.

Complete step-by-step answer:
We have $\cos \theta =\dfrac{2}{3}$
Let ABC be a right-angled triangle right-angled at A. let AB = 2 units and BC = 3units.

Hence we have $\cos B=\dfrac{2}{3}$ and $\angle B=\theta $
Now we have $B{{C}^{2}}=A{{C}^{2}}+A{{B}^{2}}$ by Pythagoras theorem.
Hence we have $9=A{{C}^{2}}+4$
Subtracting 4 from both sides, we get
$5=A{{C}^{2}}$
Hence we have $AC=\pm \sqrt{5}$, where the -ve sign just determines the quadrant in which the angle B lies.
Now we have $\tan B=\dfrac{AC}{AB}=\dfrac{\pm \sqrt{5}}{2}$ and $\sec B=\dfrac{BC}{AB}=\dfrac{3}{2}$
Hence we have $2{{\sec }^{2}}\theta +2{{\tan }^{2}}\theta -7=2{{\left( \dfrac{3}{2} \right)}^{2}}+2{{\left( \dfrac{\pm \sqrt{5}}{2} \right)}^{2}}-7$
Using ${{\left( \dfrac{a}{b} \right)}^{n}}=\dfrac{{{a}^{n}}}{{{b}^{n}}}$, we get
$2{{\sec }^{2}}\theta +2{{\tan }^{2}}\theta -7=2\times \dfrac{9}{4}+2\times \dfrac{5}{4}-7=\dfrac{9}{2}+\dfrac{5}{2}-7=7-7=0$
Hence we have $2{{\sec }^{2}}\theta +2{{\tan }^{2}}\theta -7=0$
Hence option [b] is correct.

Note: [1] Alternative solution:
We know that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$
Hence we have ${{\sin }^{2}}\theta +{{\left( \dfrac{2}{3} \right)}^{2}}=1\Rightarrow {{\sin }^{2}}\theta =\dfrac{5}{9}$
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
Hence we have ${{\tan }^{2}}\theta =\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }$
Substituting the value of ${{\sin }^{2}}\theta $ and ${{\cos }^{2}}\theta $, we get
${{\tan }^{2}}\theta =\dfrac{5}{9}\times \dfrac{9}{4}=\dfrac{5}{4}$
Now we know that ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta =\dfrac{5}{4}+1=\dfrac{9}{4}$
Hence we have $2{{\sec }^{2}}\theta +2{{\tan }^{2}}\theta -7=2\times \dfrac{9}{4}+2\times \dfrac{5}{4}-7=0$, which is the same as obtained above.
Hence option [b] is correct.
[2] In these types of questions, care should be taken about the sign of each trigonometric ratio. We should not just take one sign and reject the other without reason. This can lead to incorrect results.