Question

If $\tan \theta =2$ ,find the values of other trigonometric ratios.

Answer 1

Hint: To find the other trigonometric ratios, we have to use the trigonometric formulas. From $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $ , we can find the value of $\sec \theta $ . Using $\cos \theta =\dfrac{1}{\sec \theta }$ and $\cot \theta =\dfrac{1}{\tan \theta }$ , we can find the value of $\cos \theta $ and $\cot \theta $ . Using the formula ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ and $\csc \theta =\dfrac{1}{\sin \theta }$ , we can find the value of $\sin \theta $ and $\csc \theta $ .

Complete step by step solution:
We are given that $\tan \theta =2$ . We have to find the other trigonometric ratios. We know that
$1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $
Let us substitute the given value in the above formula.
$\begin{align}
  & \Rightarrow {{\sec }^{2}}\theta =1+{{2}^{2}}=1+4 \\
 & \Rightarrow {{\sec }^{2}}\theta =5 \\
\end{align}$
Now, we have to take the square root of the above equation.
$\Rightarrow \sec \theta =\sqrt{5}...\left( i \right)$
Now, we know that
$\cos \theta =\dfrac{1}{\sec \theta }$
Let us substitute (i) in the above formula.
$\Rightarrow \cos \theta =\dfrac{1}{\sqrt{5}}...\left( ii \right)$
We know that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ . We have to find $\sin \theta $ using this formula. Let us take ${{\cos }^{2}}\theta $ to the RHS.
$\Rightarrow {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $
Now, we have to substitute (ii) in the above formula.
$\begin{align}
  & \Rightarrow {{\sin }^{2}}\theta =1-{{\left( \dfrac{1}{\sqrt{5}} \right)}^{2}} \\
 & \Rightarrow {{\sin }^{2}}\theta =1-\dfrac{1}{5} \\
\end{align}$
Let us take the LCM of the denominators of RHS and simplify.
$\begin{align}
  & \Rightarrow {{\sin }^{2}}\theta =\dfrac{5}{5}-\dfrac{1}{5} \\
 & \Rightarrow {{\sin }^{2}}\theta =\dfrac{5-1}{5} \\
 & \Rightarrow {{\sin }^{2}}\theta =\dfrac{4}{5} \\
\end{align}$
Now, we have to take the square root of both the sides.
$\begin{align}
  & \Rightarrow \sin \theta =\sqrt{\dfrac{4}{5}} \\
 & \Rightarrow \sin \theta =\dfrac{2}{\sqrt{5}}...\left( iii \right) \\
\end{align}$
We know that $\csc \theta =\dfrac{1}{\sin \theta }$ . Let us substitute (iii) in this formula.
$\begin{align}
  & \Rightarrow \csc \theta =\dfrac{1}{\dfrac{2}{\sqrt{5}}} \\
 & \Rightarrow \csc \theta =\dfrac{\sqrt{5}}{2} \\
\end{align}$
We know that $\cot \theta =\dfrac{1}{\tan \theta }$ . Let us substitute the given value of $\tan \theta $ in this formula.
$\Rightarrow \cot \theta =\dfrac{1}{2}$
Therefore, $\sin \theta =\dfrac{2}{\sqrt{5}},\cos \theta =\dfrac{1}{\sqrt{5}},\csc \theta =\dfrac{\sqrt{5}}{2},\sec \theta =\sqrt{5},\cot \theta =\dfrac{1}{2}$ .

Note: Students must be very thorough with the formulas of trigonometric functions. They can also use other related formulas to find the values of the required trigonometric functions. For example, they can also use ${{\csc }^{2}}\theta =1+{{\cot }^{2}}\theta $ to find the value of $\csc \theta $ and then find $\sin \theta $ and so on.