Question

If the value of \[\tan \theta =\dfrac{1}{\sqrt{7}}\] , show that \[\dfrac{\cos e{{c}^{2}}\theta -{{\sec }^{2}}\theta }{cose{{c}^{2}}\theta +{{\sec }^{2}}\theta }=\dfrac{3}{4}\] .

Answer 1

Hint:In the given expression, we don’t have any \[\tan \theta \] term. For finding the value of the given expression, we have to make the expression in the form of \[\tan \theta \]. For that take \[{{\operatorname{cosec}}^{2}}\theta \] as common in both numerator and denominator. Write \[\sec \theta \] and \[\operatorname{cosec}\theta \] in terms of \[\sin \theta \] and \[\cos \theta \] . Put the value of \[\tan \theta \] and solve it further.

Complete step-by-step answer:
Let us proceed with the LHS of the given expression.
In LHS we have, \[\dfrac{\cos e{{c}^{2}}\theta -{{\sec }^{2}}\theta }{cose{{c}^{2}}\theta +{{\sec }^{2}}\theta }\]……………(1)
\[{{\operatorname{cosec}}^{2}}\theta \] is given in the numerator as well as the denominator.
Taking \[{{\operatorname{cosec}}^{2}}\theta \] common in the numerator as well as the denominator.
Solving equation (1), we get
\[\begin{align}
  & \dfrac{\cos e{{c}^{2}}\theta -{{\sec }^{2}}\theta }{cose{{c}^{2}}\theta +{{\sec }^{2}}\theta } \\
 & =\dfrac{\cos e{{c}^{2}}\theta \left( 1-\dfrac{{{\sec }^{2}}\theta }{\cos e{{c}^{2}}\theta } \right)}{\cos e{{c}^{2}}\theta \left( 1+\dfrac{{{\sec }^{2}}\theta }{\cos e{{c}^{2}}\theta } \right)} \\
\end{align}\]
\[=\dfrac{1-\dfrac{{{\sec }^{2}}\theta }{\cos e{{c}^{2}}\theta }}{1+\dfrac{{{\sec }^{2}}\theta }{\cos e{{c}^{2}}\theta }}\]…………(2)
We know that, \[\sec \theta =\dfrac{1}{\cos \theta }\]………..(3)
We also know that, \[cosec\theta =\dfrac{1}{sin\theta }\]…………………(4)
Using equation (3) and equation (4), we can write equation (2) as
\[\dfrac{1-\dfrac{\dfrac{1}{{{\cos }^{2}}\theta }}{\dfrac{1}{{{\sin }^{2}}\theta }}}{1+\dfrac{\dfrac{1}{{{\cos }^{2}}\theta }}{\dfrac{1}{{{\sin }^{2}}\theta }}}\]
On solving we get,
\[=\dfrac{1-\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }}{1+\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }}\]…………………..(5)
We know that, \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]…………(6)
Using equation (6), we can write equation (5) as \[\dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }\]……………(7)
According to the question, the value of tan θ is given.
\[\tan \theta =\dfrac{1}{\sqrt{7}}\]
Putting the value tan θ in equation (7), we get
\[\begin{align}
  & \dfrac{1-{{\left( \dfrac{1}{\sqrt{7}} \right)}^{2}}}{1+{{\left( \dfrac{1}{\sqrt{7}} \right)}^{2}}} \\
 & =\dfrac{\dfrac{7-1}{7}}{\dfrac{7+1}{7}} \\
 & =\dfrac{6}{8} \\
 & =\dfrac{3}{4} \\
\end{align}\]
So, \[\dfrac{\cos e{{c}^{2}}\theta -{{\sec }^{2}}\theta }{cose{{c}^{2}}\theta +{{\sec }^{2}}\theta }=\dfrac{3}{4}\].
Therefore, LHS=RHS.
Hence, proved.

Note: This question can also be solved by putting the value of \[\operatorname{cosec}\theta \] and \[\sec \theta \] .But the values of \[\operatorname{cosec}\theta \] and \[\sec \theta \] is not given in the question. So, by using the value of \[\tan \theta \] , we can find the value of \[\operatorname{cosec}\theta \] and \[\sec \theta \] .

\[tan\theta =\dfrac{1}{\sqrt{7}}\]
Using Pythagora's theorem, we can find hypotenuses.
Hypotenuse= \[\sqrt{{{\left( height \right)}^{2}}+{{\left( base \right)}^{2}}}\]
\[\begin{align}
  & =\sqrt{{{\left( 1 \right)}^{2}}+{{\left( \sqrt{7} \right)}^{2}}} \\
 & =\sqrt{1+7} \\
 & =\sqrt{8} \\
\end{align}\]
\[\operatorname{cosec}\theta =\dfrac{\sqrt{8}}{1}\] and \[\sec \theta =\dfrac{\sqrt{8}}{\sqrt{7}}\] .
And after putting the value of \[\operatorname{cosec}\theta \] and \[\sec \theta \] in the given expression, we can get our required answer.