Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

If $\theta $ is an acute angle and $\sin \theta = \cos \theta $, find the value of $2{\left( {\tan \theta } \right)^2} + {\left( {\sin \theta } \right)^2} - 1$.

seo-qna
Last updated date: 17th Jul 2024
Total views: 450.6k
Views today: 10.50k
Answer
VerifiedVerified
450.6k+ views
Hint: Here, we will be finding the value of angle $\theta $ from the given equation and then we will be using the values like $\tan {45^0} = 1$ and \[\sin {45^0} = \dfrac{1}{{\sqrt 2 }}\] given in the trigonometric table in order to obtain the value of the given expression.

Complete step-by-step answer:
Given, $\sin \theta = \cos \theta $ where $\theta $ is an acute angle
As we know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$

The given equation can be rearranged as $
   \Rightarrow \dfrac{{\sin \theta }}{{\cos \theta }} = 1 \\
   \Rightarrow \tan \theta = 1{\text{ }} \to {\text{(1)}} \\
 $

Also we know that tangent of 45 degrees is equal to 1 i.e., $\tan {45^0} = 1{\text{ }} \to {\text{(2)}}$

By comparing equations (1) and (2), we will get the value for $\theta $
$ \Rightarrow \theta = {45^0}$

Here, we have considered only $\theta = {45^0}$ because it is given that $\theta $ is an acute angle (angle which is less than 90 degrees).

Let us suppose the value of expression whose value we need to find is x
So, $x = 2{\left( {\tan \theta } \right)^2} + {\left( {\sin \theta } \right)^2} - 1$

Now, let us substitute the value of $\theta = {45^0}$ in the above expression in order to find the value of x.
\[
   \Rightarrow x = 2{\left( {\tan \theta } \right)^2} + {\left( {\sin \theta } \right)^2} - 1 \\
   \Rightarrow x = 2{\left( {\tan {{45}^0}} \right)^2} + {\left( {\sin {{45}^0}} \right)^2} - 1{\text{ }} \to {\text{(3)}} \\
 \]

According to trigonometric table, we can write
\[\tan {45^0} = 1\] and \[\sin {45^0} = \dfrac{1}{{\sqrt 2 }}\]

Putting these values in equation (3), we get
\[ \Rightarrow x = 2{\left( 1 \right)^2} + {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2} - 1 = 2 + \dfrac{1}{2} - 1 = 1 + \dfrac{1}{2} = \dfrac{{2 + 1}}{2} = \dfrac{3}{2}\]

Therefore, the value of the expression is given by \[2{\left( {\tan \theta } \right)^2} + {\left( {\sin \theta } \right)^2} - 1 = \dfrac{3}{2}\].

Note: In this problem, the important step lies in the determination of the angle $\theta $ because $\tan \theta = 1$ gives various values of $\theta $ as $\theta = {45^0},{225^0},{405^0}$, etc but in the problem it is given that $\theta $ is an acute angle so we will consider only that value of $\theta $ which measures less than ${90^0}$. That’s why the only possible result of $\tan \theta = 1$ is $\theta = {45^0}$.