Answer
Verified
400.5k+ views
Hint: 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:
$\begin{align}
& 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \\
& \Rightarrow {{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1 \\
\end{align}$
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.
$\begin{align}
& \dfrac{1}{{{\cos }^{2}}\theta }=3 \\
& \Rightarrow {{\cos }^{2}}\theta =\dfrac{1}{3} \\
\end{align}$
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,
$\begin{align}
& {{\sin }^{2}}\theta =1-\dfrac{1}{3} \\
& \Rightarrow {{\sin }^{2}}\theta =\dfrac{2}{3} \\
\end{align}$
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,
$\begin{align}
& \dfrac{2-\cos e{{c}^{2}}\theta }{2+\cos e{{c}^{2}}\theta } \\
& =\dfrac{2-\dfrac{3}{2}}{2+\dfrac{3}{2}} \\
& =\dfrac{1}{7} \\
\end{align}$
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 $.
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:
$\begin{align}
& 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \\
& \Rightarrow {{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1 \\
\end{align}$
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.
$\begin{align}
& \dfrac{1}{{{\cos }^{2}}\theta }=3 \\
& \Rightarrow {{\cos }^{2}}\theta =\dfrac{1}{3} \\
\end{align}$
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,
$\begin{align}
& {{\sin }^{2}}\theta =1-\dfrac{1}{3} \\
& \Rightarrow {{\sin }^{2}}\theta =\dfrac{2}{3} \\
\end{align}$
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,
$\begin{align}
& \dfrac{2-\cos e{{c}^{2}}\theta }{2+\cos e{{c}^{2}}\theta } \\
& =\dfrac{2-\dfrac{3}{2}}{2+\dfrac{3}{2}} \\
& =\dfrac{1}{7} \\
\end{align}$
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:
$\begin{align}
& {{H}^{2}}={{B}^{2}}+{{P}^{2}} \\
& 3=1+{{P}^{2}} \\
& \Rightarrow {{P}^{2}}=2 \\
& \Rightarrow P=\sqrt{2} \\
\end{align}$
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,
$\begin{align}
& \dfrac{{{\tan }^{2}}\theta -\cos e{{c}^{2}}\theta }{{{\tan }^{2}}\theta +\cos e{{c}^{2}}\theta } \\
& =\dfrac{{{\left( \sqrt{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{3}}{\sqrt{2}} \right)}^{2}}}{{{\left( \sqrt{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{\sqrt{2}} \right)}^{2}}} \\
& =\dfrac{2-\dfrac{3}{2}}{2+\dfrac{3}{2}} \\
& =\dfrac{1}{7} \\
\end{align}$
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.
$\sec \theta =\dfrac{H}{B}$
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:
$\begin{align}
& {{H}^{2}}={{B}^{2}}+{{P}^{2}} \\
& 3=1+{{P}^{2}} \\
& \Rightarrow {{P}^{2}}=2 \\
& \Rightarrow P=\sqrt{2} \\
\end{align}$
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,
$\begin{align}
& \dfrac{{{\tan }^{2}}\theta -\cos e{{c}^{2}}\theta }{{{\tan }^{2}}\theta +\cos e{{c}^{2}}\theta } \\
& =\dfrac{{{\left( \sqrt{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{3}}{\sqrt{2}} \right)}^{2}}}{{{\left( \sqrt{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{\sqrt{2}} \right)}^{2}}} \\
& =\dfrac{2-\dfrac{3}{2}}{2+\dfrac{3}{2}} \\
& =\dfrac{1}{7} \\
\end{align}$
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.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Write an application to the principal requesting five class 10 english CBSE
Difference Between Plant Cell and Animal Cell
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
List some examples of Rabi and Kharif crops class 8 biology CBSE