Question

If $\cot \theta =\dfrac{3}{4}$, then prove that the expression $\sqrt{\dfrac{\sec \theta -\operatorname{cosec}\theta }{\sec \theta +\operatorname{cosec}\theta }}=\dfrac{1}{\sqrt{7}}$.

Answer 1

Hint:In order to solve this question, we should know that $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$. And therefore, to use this formula, we will start our question from the equality which we have to prove. We need to also remember that $\sec \theta ,\operatorname{cosec}\theta $ can be expressed as $\dfrac{1}{\cos \theta },\dfrac{1}{\sin \theta }$ respectively.

Complete step-by-step answer:
In this question, we have been asked to prove that $\sqrt{\dfrac{\sec \theta -\operatorname{cosec}\theta }{\sec \theta +\operatorname{cosec}\theta }}=\dfrac{1}{\sqrt{7}}$, when we are given $\cot \theta =\dfrac{3}{4}$. So, to prove this equality, we will consider the left hand side or the LHS of the equality,
$LHS=\sqrt{\dfrac{\sec \theta -\operatorname{cosec}\theta }{\sec \theta +\operatorname{cosec}\theta }}$
We know that $\sec \theta ,\operatorname{cosec}\theta $ can be written as $\dfrac{1}{\cos \theta },\dfrac{1}{\sin \theta }$ respectively. So, applying that, we get,
$LHS=\sqrt{\dfrac{\dfrac{1}{\cos \theta }-\dfrac{1}{\sin \theta }}{\dfrac{1}{\cos \theta }+\dfrac{1}{\sin \theta }}}$
Now, we will take the LCM of the terms of the numerator and the terms of the denominator, So, we get the above equation as,
$LHS=\sqrt{\dfrac{\dfrac{\left( \sin \theta -\cos \theta \right)}{\cos \theta \sin \theta }}{\dfrac{\left( \sin \theta +\cos \theta \right)}{\cos \theta \sin \theta }}}$
Simplifying it further, we get the LHS as,
$LHS=\sqrt{\dfrac{\left( \sin \theta -\cos \theta \right)\left( \sin \theta \cos \theta \right)}{\left( \sin \theta +\cos \theta \right)\left( \sin \theta \cos \theta \right)}}$
Now, we know that the common terms in the numerator and the denominator gets cancelled. So, we can write the LHS as,
$LHS=\sqrt{\dfrac{\sin \theta -\cos \theta }{\sin \theta +\cos \theta }}$
Now, we will take $\sin \theta $ as common from the numerator and denominator of LHS, so we get,
$\begin{align}
  & LHS=\sqrt{\dfrac{\sin \theta \left( 1-\dfrac{\cos \theta }{\sin \theta } \right)}{\sin \theta \left( 1+\dfrac{\cos \theta }{\sin \theta } \right)}} \\
 & \Rightarrow LHS=\sqrt{\dfrac{1-\left( \dfrac{\cos \theta }{\sin \theta } \right)}{1+\left( \dfrac{\cos \theta }{\sin \theta } \right)}} \\
\end{align}$
We know that $\dfrac{\cos \theta }{\sin \theta }$ can be expressed as $\cot \theta $. So, applying the same, we get the LHS as,
$LHS=\sqrt{\dfrac{1-\cot \theta }{1+\cot \theta }}$
No, we have been given in the question that $\cot \theta =\dfrac{3}{4}$. So, we will substitute the value and get,
$LHS=\sqrt{\dfrac{1-\dfrac{3}{4}}{1+\dfrac{3}{4}}}$
We will now take the LCM of both the terms of the numerator and the denominator. So, we get,
$\begin{align}
  & LHS=\sqrt{\dfrac{\dfrac{4-3}{4}}{\dfrac{4+3}{4}}} \\
 & \Rightarrow LHS=\sqrt{\dfrac{1\times 4}{7\times 4}} \\
 & \Rightarrow LHS=\sqrt{\dfrac{1}{7}} \\
 & \Rightarrow LHS=\dfrac{1}{\sqrt{7}} \\
\end{align}$
Which is the same as the right hand side of the given expression, so LHS = RHS.
Hence, we have proved the expression given in the question.

Note: We can also solve this question by taking out $\sec \theta $ as common from the numerator and the denominator. We know that $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$, and $\cos \theta $ can be written as $\dfrac{1}{\sec \theta }$ and $\sin \theta $ can be written as $\dfrac{1}{\operatorname{cosec}\theta }$. Therefore, we will get $\cot \theta =\dfrac{\dfrac{1}{\sec \theta }}{\dfrac{1}{\operatorname{cosec}\theta }}$ , which is same as $\cot \theta =\dfrac{\operatorname{cosec}\theta }{\sec \theta }$.Substitute the value in given expression and get the required answer.