Question

If θ is an acute angle such that ${{\sec }^{2}}\theta =3$ , then the value of $\dfrac{{{\tan }^{2}}\theta -\cos e{{c}^{2}}\theta }{{{\tan }^{2}}\theta +\cos e{{c}^{2}}\theta }$ is:
A. $\dfrac{4}{7}$
B. $\dfrac{3}{7}$
C. $\dfrac{2}{7}$
D. $\dfrac{1}{7}$

Answer 1

The value of expression that we need to find contains ${{\tan }^{2}}\theta$ in the numerator and the denominator so we can write ${{\tan }^{2}}\theta$ as ${{\sec }^{2}}\theta -1$ in that expression. And substitute the value of ${{\sec }^{2}}\theta$ given in the question. Now, we are left with $\cos e{{c}^{2}}\theta$ which can be calculated from ${{\sec }^{2}}\theta$ .

Complete step-by-step answer:
The expression given in the question is:
$\dfrac{{{\tan }^{2}}\theta -\cos e{{c}^{2}}\theta }{{{\tan }^{2}}\theta +\cos e{{c}^{2}}\theta }$
From the trigonometric identities we know that:
$\begin{aligned} & 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \\\ & \Rightarrow {{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1 \\\ \end{aligned}$
Substituting the value of ${{\tan }^{2}}\theta$ that we have calculated above in the given expression we get,
$\dfrac{{{\sec }^{2}}\theta -1-\cos e{{c}^{2}}\theta }{{{\sec }^{2}}\theta -1+\cos e{{c}^{2}}\theta }$
The value of ${{\sec }^{2}}\theta =3$ is given in the question so plugging this value in the above equation we get,
$\dfrac{3-1-\cos e{{c}^{2}}\theta }{3-1+\cos e{{c}^{2}}\theta }$
$=\dfrac{2-\cos e{{c}^{2}}\theta }{2+\cos e{{c}^{2}}\theta }$ ……… Eq. (1)
We can find the value of $\cos e{{c}^{2}}\theta$ from ${{\sec }^{2}}\theta$ as follows:
${{\sec }^{2}}\theta =3$
We know that ${{\sec }^{2}}\theta =\dfrac{1}{{{\cos }^{2}}\theta }$ so rewriting the above equation in the following manner.
$\begin{aligned} & \dfrac{1}{{{\cos }^{2}}\theta }=3 \\\ & \Rightarrow {{\cos }^{2}}\theta =\dfrac{1}{3} \\\ \end{aligned}$
From the trigonometric identities we know that,
${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta$
Substituting the value of ${{\cos }^{2}}\theta$ as $\dfrac{1}{3}$ in the above equation we get,
$\begin{aligned} & {{\sin }^{2}}\theta =1-\dfrac{1}{3} \\\ & \Rightarrow {{\sin }^{2}}\theta =\dfrac{2}{3} \\\ \end{aligned}$
From the trigonometric ratios we know that $\cos e{{c}^{2}}\theta$ is the inverse of ${{\sin }^{2}}\theta$ so we can write $\cos e{{c}^{2}}\theta$ as:
$\cos e{{c}^{2}}\theta =\dfrac{3}{2}$
Now, substituting this value of $\cos e{{c}^{2}}\theta$ θ in eq. (1) we get,
$\begin{aligned} & \dfrac{2-\cos e{{c}^{2}}\theta }{2+\cos e{{c}^{2}}\theta } \\\ & =\dfrac{2-\dfrac{3}{2}}{2+\dfrac{3}{2}} \\\ & =\dfrac{1}{7} \\\ \end{aligned}$
From the above calculation, we have calculated the value of the given expression as $\dfrac{1}{7}$ .
Hence, the correct option is (d).

Note: The other way of solving the above problem is as follows:
It is given that ${{\sec }^{2}}\theta =3$ so taking square root on both the sides we get,
$\sec \theta =\sqrt{3}$
Here, we are taking the positive value because θ is given as acute angle.
In the below figure, we have drawn a triangle ABC right angled at B and the figure is also showing an angle $\theta$.

We know that from the trigonometric ratios that:
$\sec \theta =\dfrac{H}{B}$
In the above formula, H stands for hypotenuse of the triangle with respect to angle $\theta$ and B stands for the base of the triangle with respect to angle $\theta$.
$\sqrt{3}=\dfrac{H}{B}$
From the above formula, we can calculate the perpendicular (or the other side of the triangle) by Pythagoras theorem as follows:
$\begin{aligned} & {{H}^{2}}={{B}^{2}}+{{P}^{2}} \\\ & 3=1+{{P}^{2}} \\\ & \Rightarrow {{P}^{2}}=2 \\\ & \Rightarrow P=\sqrt{2} \\\ \end{aligned}$
In the above calculation, P stands for perpendicular of the triangle.
Now, we have P, B and H. We can calculate the value of $\tan \theta$ and $\cos ec\theta$ .
$\tan \theta =\dfrac{P}{B}$
Plugging the values of “P” and “B” in the above equation we get,
$\tan \theta =\sqrt{2}$
$\cos ec\theta =\dfrac{H}{P}$
Plugging the values of “P” and “H” in the above equation we get,
$\cos ec\theta =\dfrac{\sqrt{3}}{\sqrt{2}}$
Substituting the above values in this expression $\dfrac{{{\tan }^{2}}\theta -\cos e{{c}^{2}}\theta }{{{\tan }^{2}}\theta +\cos e{{c}^{2}}\theta }$ we get,
$\begin{aligned} & \dfrac{{{\tan }^{2}}\theta -\cos e{{c}^{2}}\theta }{{{\tan }^{2}}\theta +\cos e{{c}^{2}}\theta } \\\ & =\dfrac{{{\left( \sqrt{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{3}}{\sqrt{2}} \right)}^{2}}}{{{\left( \sqrt{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{\sqrt{2}} \right)}^{2}}} \\\ & =\dfrac{2-\dfrac{3}{2}}{2+\dfrac{3}{2}} \\\ & =\dfrac{1}{7} \\\ \end{aligned}$
Hence, we are getting the value of the given expression is $\dfrac{1}{7}$ which is the same as we have calculated in the solution part.