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# If $\tan \theta =\dfrac{1}{\sqrt{7}}$, show that:$\dfrac{\text{cose}{{\text{c}}^{2}}\theta -{{\sec }^{2}}\theta }{\text{cose}{{\text{c}}^{2}}\theta +{{\sec }^{2}}\theta }=\dfrac{3}{4}$

Last updated date: 15th Sep 2024
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Hint: In this question, from the given values of tan function by using the trigonometric identities we can find the values of the sec and cosec functions. Then on substituting the respective values in the given expression of the question we can calculate the left hand side value and the right hand side value. Then on comparing the values obtained, we get the result.
\begin{align} & \tan \theta =\dfrac{\sin \theta }{\cos \theta } \\ & {{\sec }^{2}}\theta -1={{\tan }^{2}}\theta \\ & \text{sec}\theta =\dfrac{1}{\cos \theta } \\ & \text{cosec}\theta =\dfrac{1}{\sin \theta } \\ \end{align}

Complete step by step answer:
Now, from the given question we have
$\tan \theta =\dfrac{1}{\sqrt{7}}\ \ \ \ \ ...(a)$
Now, by using the trigonometric identity which gives the relation between the function that are mentioned in the hint, we get the following
$\Rightarrow \tan \theta =\dfrac{\sin \theta }{\cos \theta }$
Now, this can also be written as the following using the other relations given in the hint as follow
\begin{align} & \Rightarrow {{\sec }^{2}}\theta -1={{\tan }^{2}}\theta \\ & \Rightarrow {{\left( \dfrac{1}{\cos \theta } \right)}^{2}}-1={{\tan }^{2}}\theta \\ & \Rightarrow {{\left( \dfrac{1}{\cos \theta } \right)}^{2}}={{\tan }^{2}}\theta +1 \\ \end{align}
Let us now substitute the value from the question and as well as from equation (a) in this
\begin{align} & \Rightarrow {{\left( \dfrac{1}{\cos \theta } \right)}^{2}}={{\left( \dfrac{1}{\sqrt{7}} \right)}^{2}}+1 \\ & \Rightarrow {{\left( \dfrac{1}{\cos \theta } \right)}^{2}}=\left( \dfrac{1}{7} \right)+1 \\ & \Rightarrow {{\left( \dfrac{1}{\cos \theta } \right)}^{2}}=\dfrac{8}{7} \\ \end{align}
Now, this can be further written as
\begin{align} & \Rightarrow \left( \dfrac{1}{\cos \theta } \right)=\sqrt{\dfrac{8}{7}} \\ & \Rightarrow \cos \theta =\sqrt{\dfrac{7}{8}} \\ \end{align}
Hence, we can get the following from the relation given in the hint as follows
\begin{align} & \Rightarrow \cos \theta =\sqrt{\dfrac{7}{8}} \\ & \left( \sec \theta =\dfrac{1}{\cos \theta } \right) \\ & \Rightarrow \sec \theta =\dfrac{1}{\sqrt{\dfrac{7}{8}}} \\ & \Rightarrow \sec \theta =\sqrt{\dfrac{8}{7}} \\ \end{align}
Now, using the relation between the sin and cos function, we have
$\Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
Now, this can be used to get the expression which can be written as
\begin{align} & \Rightarrow {{\left( \sqrt{\dfrac{7}{8}} \right)}^{2}}+{{\sin }^{2}}\theta =1 \\ & \Rightarrow \dfrac{7}{8}+{{\sin }^{2}}\theta =1 \\ & \Rightarrow {{\sin }^{2}}\theta =1-\dfrac{7}{8} \\ & \Rightarrow {{\sin }^{2}}\theta =\dfrac{1}{8} \\ & \Rightarrow \sin \theta =\sqrt{\dfrac{1}{8}} \\ \end{align}
Hence, we can get the following from the relation given in the hint as follows
\begin{align} & \Rightarrow \sin \theta =\sqrt{\dfrac{1}{8}} \\ & \left( \text{cosec}\theta =\dfrac{1}{\sin \theta } \right) \\ & \Rightarrow \text{cosec}\theta =\dfrac{1}{\sqrt{\dfrac{1}{8}}} \\ & \Rightarrow \text{cosec}\theta =\sqrt{\dfrac{8}{1}}=\sqrt{8} \\ \end{align}
Now, from the given expression in the question, on substituting the values, we have
$\dfrac{\text{cose}{{\text{c}}^{2}}\theta -{{\sec }^{2}}\theta }{\text{cose}{{\text{c}}^{2}}\theta +{{\sec }^{2}}\theta }=\dfrac{3}{4}$
Let us first consider the left hand side and calculate its value
$L.H.S=\dfrac{\text{cose}{{\text{c}}^{2}}\theta -{{\sec }^{2}}\theta }{\text{cose}{{\text{c}}^{2}}\theta +{{\sec }^{2}}\theta }$
Thus, the value of right hand side is equal to left hand side
Hence, it is verified that
$\dfrac{\text{cose}{{\text{c}}^{2}}\theta -{{\sec }^{2}}\theta }{\text{cose}{{\text{c}}^{2}}\theta +{{\sec }^{2}}\theta }=\dfrac{3}{4}$
Substituting the value of $\text{cosec}\theta =\sqrt{8}$ and $\sec \theta =\sqrt{\dfrac{8}{7}}$ in the above equation we get,

\begin{align} & L.H.S=\dfrac{{{\left( \sqrt{8} \right)}^{2}}-{{\left( \sqrt{\dfrac{8}{7}} \right)}^{2}}}{{{\left( \sqrt{8} \right)}^{2}}+{{\left( \sqrt{\dfrac{8}{7}} \right)}^{2}}} \\ & L.H.S=\dfrac{8-\dfrac{8}{7}}{8+\dfrac{8}{7}} \\ \end{align}
Taking 7 as L.C.M in both the numerator and the denominator we get,
\begin{align} & L.H.S=\dfrac{\dfrac{56-8}{7}}{\dfrac{56+8}{7}} \\ & L.H.S=\dfrac{\dfrac{48}{7}}{\dfrac{64}{7}} \\ & L.H.S=\dfrac{48}{64} \\ & L.H.S=\dfrac{3}{4} \\ \end{align}

Note: The other way of solving the above problem is as follows:
$\dfrac{\text{cose}{{\text{c}}^{2}}\theta -{{\sec }^{2}}\theta }{\text{cose}{{\text{c}}^{2}}\theta +{{\sec }^{2}}\theta }=\dfrac{3}{4}$
We are going to solve the L.H.S of the above equation.
$\dfrac{\text{cose}{{\text{c}}^{2}}\theta -{{\sec }^{2}}\theta }{\text{cose}{{\text{c}}^{2}}\theta +{{\sec }^{2}}\theta }$……….. Eq. (b)
We know the following trigonometric properties:
\begin{align} & \text{cosec}\theta =\dfrac{1}{\sin \theta } \\ & \sec \theta =\dfrac{1}{\cos \theta } \\ \end{align}
Using the above properties in eq. (b) we get,
$\dfrac{\dfrac{1}{{{\sin }^{2}}\theta }-\dfrac{1}{{{\cos }^{2}}\theta }}{\dfrac{1}{{{\sin }^{2}}\theta }+\dfrac{1}{{{\cos }^{2}}\theta }}$
Taking ${{\sin }^{2}}\theta {{\cos }^{2}}\theta$ as common in the numerator and denominator we get,
\begin{align} & \dfrac{\dfrac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }}{\dfrac{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }} \\ & =\dfrac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta } \\ \end{align}
There are following trigonometric identities which can be used to simplify the above expression are as follows:
\begin{align} & {{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta \\ & {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \\ \end{align}
\begin{align} & \dfrac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta } \\ & =\cos 2\theta \\ \end{align}
It is also given in the question that $\tan \theta =\dfrac{1}{\sqrt{7}}$ and we know that:
$\cos 2\theta =\dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }$
Substituting the above expansion in the above result we get,
\begin{align} & \cos 2\theta \\ & =\dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \\ \end{align}
Substituting $\tan \theta =\dfrac{1}{\sqrt{7}}$ in the above equation we get,
$\dfrac{1-{{\left( \dfrac{1}{\sqrt{7}} \right)}^{2}}}{1+{{\left( \dfrac{1}{\sqrt{7}} \right)}^{2}}}$
\begin{align} & =\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}} \\ & =\dfrac{\dfrac{6}{7}}{\dfrac{8}{7}}=\dfrac{3}{4} \\ \end{align}
R.H.S of the given equation is given as $\dfrac{3}{4}$. Hence, we have proved the given equation.