Question

If $\tan \theta =\dfrac{1}{\sqrt{5}}$ and $\theta $ lies in the first quadrant, then find the value of $\cos \theta $.
A. $\dfrac{1}{\sqrt{6}}$
B. $\dfrac{\sqrt{5}}{\sqrt{6}}$
C. $\dfrac{-1}{\sqrt{6}}$
D. $\dfrac{-\sqrt{5}}{\sqrt{6}}$

Answer 1

Hint: We have the value of $\tan \theta $ and the $\theta$ lies in the first quadrant. Now we will assume a triangle $ABC$ and calculate the value of $\tan \theta $ and equate it to the given value. From the Pythagoras theorem we can calculate the length of the other side and then we will have all the lengths of the sides of the triangle. For this we can calculate any trigonometric ratio and after finding the ratios, according to the quadrant that $\theta $ lies, we will change the sign of the trigonometric ratio.

Complete step by step answer:
Given that, $\tan \theta =\dfrac{1}{\sqrt{5}}$ and $\theta $ lies in first quadrant.
Let us assume a triangle $ABC$ as shown in figure.

In the above triangle the value of $\tan \theta $ is given by calculating the ratio of the opposite side to $\theta$ and the adjacent side to $\theta$, mathematically
$\begin{align}
  & \tan \theta =\dfrac{\text{opposite side to }\theta }{\text{Adjacent side to }\theta } \\
 & \Rightarrow \tan \theta =\dfrac{BA}{BC} \\
 & \Rightarrow \tan \theta =\dfrac{c}{a} \\
\end{align}$
But we have the value of $\tan \theta $ equal to $\dfrac{1}{\sqrt{5}}$.
$\begin{align}
  & \therefore \tan \theta =\dfrac{1}{\sqrt{5}} \\
 & \Rightarrow \dfrac{c}{a}=\dfrac{1}{\sqrt{5}} \\
 & \Rightarrow c:a=1:\sqrt{5} \\
\end{align}$
We can write the values of $c$ and $a$ as $c=1\times x=x$, $a=\sqrt{5}\times x=\sqrt{5}x$.
In the triangle $ABC$, using Pythagoras theorem
${{b}^{2}}={{a}^{2}}+{{c}^{2}}$
Substituting the values of $c$ and $a$ in the above equation, then we will get
$\begin{align}
  & {{b}^{2}}={{\left( \sqrt{5}x \right)}^{2}}+{{x}^{2}} \\
 & \Rightarrow {{b}^{2}}=5{{x}^{2}}+{{x}^{2}} \\
 & \Rightarrow {{b}^{2}}=6{{x}^{2}} \\
 & \Rightarrow b=\sqrt{6{{x}^{2}}} \\
 & \Rightarrow b=\sqrt{6}x \\
\end{align}$
In the triangle $ABC$, we have the lengths of the sides as
$AB=c=x$
$BC=a=\sqrt{5}x$
$CD=b=\sqrt{6}x$
From the above values we can find the all trigonometric ratios. Now the value of $\cos \theta $ can be obtained by taking the ratio of adjacent side to $\theta$ and hypotenuse of the triangle.
$\begin{align}
  & \therefore \cos \theta =\dfrac{\text{adjacent side to }\theta }{\text{hypotenuse}} \\
 & \Rightarrow \cos \theta =\dfrac{CB}{CA} \\
 & \Rightarrow \cos \theta =\dfrac{a}{b} \\
 & \Rightarrow \cos \theta =\dfrac{\sqrt{5}x}{\sqrt{6}x} \\
 & \Rightarrow \cos \theta =\dfrac{\sqrt{5}}{\sqrt{6}} \\
\end{align}$
Given that the $\theta$ lies in the first quadrant, the values of all trigonometric ratios for $\theta$ lies in the first quadrant are positive.
$\therefore \cos \theta =\dfrac{\sqrt{5}}{\sqrt{6}}$

So, the correct answer is “Option B”.

Note: We can simply calculate this problem by knowing trigonometric identity. We have the trigonometric identity ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$, from this relation the value of $\sec \theta $ is given by
$\begin{align}
  & {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 \\
 & \Rightarrow {{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \\
 & \Rightarrow {{\sec }^{2}}\theta =1+{{\left( \dfrac{1}{\sqrt{5}} \right)}^{2}} \\
 & \Rightarrow {{\sec }^{2}}\theta =1+\dfrac{1}{5} \\
 & \Rightarrow {{\sec }^{2}}\theta =\dfrac{1\times 5+1}{5} \\
 & \Rightarrow {{\sec }^{2}}\theta =\dfrac{6}{5} \\
 & \Rightarrow \sec \theta =\dfrac{\sqrt{6}}{\sqrt{5}} \\
\end{align}$
We know that $\sec \theta $ is the inverse trigonometric ratio of $\cos \theta $.
$\begin{align}
  & \therefore \cos \theta =\dfrac{1}{\sec \theta } \\
 & \Rightarrow \cos \theta =\dfrac{1}{\dfrac{\sqrt{6}}{\sqrt{5}}} \\
 & \Rightarrow \cos \theta =\dfrac{\sqrt{5}}{\sqrt{6}} \\
\end{align}$
From both the methods we got the same result.