Question

Which of the below is the trigonometric function $\sec \theta $ equal to?
a) $\dfrac{1}{\sqrt{1-{{\cos }^{2}}\theta }}$
b) $\dfrac{\sqrt{1-{{\cot }^{2}}\theta }}{\cot \theta }$
c) $\dfrac{\cot \theta }{\sqrt{1-{{\cot }^{2}}\theta }}$
d) \[\dfrac{\sqrt{\text{cose}{{\text{c}}^{2}}\theta -1}}{\text{cosec}\theta }\]

Answer 1

Hint: Since we have $\sec \theta $, we need to find which of the options is equal to $\sec \theta $. Since we know that $\sec \theta =\dfrac{1}{\cos \theta }$, so we need to find which among the options equals to $\dfrac{1}{\cos \theta }$. Therefore, convert all the given options into sine and cosine and find the simplest form and compare with the value given in the question.

Complete step by step answer:
a) $\dfrac{1}{\sqrt{1-{{\cos }^{2}}\theta }}$
We have: $LHS=\dfrac{1}{\sqrt{1-{{\cos }^{2}}\theta }}......(1)$
Since we know that: ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$
So, we can write equation (1) as:
$\begin{align}
  & LHS=\dfrac{1}{\sqrt{{{\sin }^{2}}\theta }} \\
 & =\dfrac{1}{\sin \theta }......(2)
\end{align}$
As we know that $\dfrac{1}{\sin \theta }=\text{cosec}\theta $
So, we have:
$LHS=\text{cosec}\theta $
Therefore, $LHS\ne RHS$
Hence option (a) is incorrect.

b) $\dfrac{\sqrt{1-{{\cot }^{2}}\theta }}{\cot \theta }$
We have: $LHS=\dfrac{\sqrt{1-{{\cot }^{2}}\theta }}{\cot \theta }......(1)$
Since we know that: $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$
So, we can write equation (1) as:
\[\begin{align}
  & LHS=\dfrac{\sqrt{1-{{\left( \dfrac{\cos \theta }{\sin \theta } \right)}^{2}}}}{\left( \dfrac{\cos \theta }{\sin \theta } \right)} \\
 & =\dfrac{\sqrt{1-{{\cos }^{2}}\theta }}{\cos \theta }......(2)
\end{align}\]
Since we know that: ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$
So, we can write equation (2) as:
$\begin{align}
  & LHS=\dfrac{\sqrt{{{\sin }^{2}}\theta }}{\cos \theta } \\
 & =\dfrac{\sin \theta }{\cos \theta }......(3)
\end{align}$

As we know that $\dfrac{\sin \theta }{\cos \theta }=\tan \theta $
So, we have:
$LHS=\tan \theta $
Therefore, $LHS\ne RHS$
Hence option (b) is incorrect.

c) $\dfrac{\cot \theta }{\sqrt{1-{{\cot }^{2}}\theta }}$
We have: $LHS=\dfrac{\cot \theta }{\sqrt{1-{{\cot }^{2}}\theta }}......(1)$
Since we know that: $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$
So, we can write equation (1) as:
\[\begin{align}
  & LHS=\dfrac{\left( \dfrac{\cos \theta }{\sin \theta } \right)}{\sqrt{1-{{\left( \dfrac{\cos \theta }{\sin \theta } \right)}^{2}}}} \\
 & =\dfrac{\cos \theta }{\sqrt{1-{{\cos }^{2}}\theta }}......(2)
\end{align}\]
Since we know that: ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$
So, we can write equation (2) as:
$\begin{align}
  & LHS=\dfrac{\cos \theta }{\sqrt{{{\sin }^{2}}\theta }} \\
 & =\dfrac{\cos \theta }{\sin \theta }......(3)
\end{align}$

As we know that $\dfrac{\cos \theta }{\sin \theta }=\cot \theta $
So, we have:
$LHS=\cot \theta $
Therefore, $LHS\ne RHS$
Hence option (c) is incorrect.

d) \[\dfrac{\sqrt{\text{cose}{{\text{c}}^{2}}\theta -1}}{\text{cosec}\theta }\]
We have: $LHS=\dfrac{\sqrt{\text{cose}{{\text{c}}^{2}}\theta -1}}{\text{cosec}\theta }......(1)$
Since we know that: $\text{cose}{{\text{c}}^{2}}\theta =1+{{\cot }^{2}}\theta $
So, we can write equation (1) as:
$\begin{align}
  & LHS=\dfrac{\sqrt{\text{co}{{\text{t}}^{2}}\theta }}{\text{cosec}\theta } \\
 & =\dfrac{\text{cot}\theta }{\text{cosec}\theta }......(2)
\end{align}$
As we know that $\dfrac{1}{\sin \theta }=\text{cosec}\theta $ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$
So, we have:
$\begin{align}
  & LHS=\dfrac{\dfrac{\text{cos}\theta }{\sin \theta }}{\dfrac{1}{\sin \theta }} \\
 & =\text{cos}\theta
\end{align}$
Therefore, $LHS\ne RHS$
Hence option (d) is incorrect.
Note:
Always remember that whenever we are given a trigonometric function to solve, try to simplify the expression in terms of sine and cosine of the given angles. It makes the solution simpler and hence apply sine and cosine relations to solve the question.