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Question: If \(\theta \) is an acute angle and \(\sin \theta = \cos \theta \), find the value of \(2{\left( {\...

If θ\theta is an acute angle and sinθ=cosθ\sin \theta = \cos \theta , find the value of 2(tanθ)2+(sinθ)212{\left( {\tan \theta } \right)^2} + {\left( {\sin \theta } \right)^2} - 1.

Explanation

Solution

Hint: Here, we will be finding the value of angle θ\theta from the given equation and then we will be using the values like tan450=1\tan {45^0} = 1 and sin450=12\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, sinθ=cosθ\sin \theta = \cos \theta where θ\theta is an acute angle
As we know that tanθ=sinθcosθ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}

The given equation can be rearranged as sinθcosθ=1 tanθ=1 (1)  \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., tan450=1 (2)\tan {45^0} = 1{\text{ }} \to {\text{(2)}}

By comparing equations (1) and (2), we will get the value for θ\theta
θ=450\Rightarrow \theta = {45^0}

Here, we have considered only θ=450\theta = {45^0} because it is given that θ\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(tanθ)2+(sinθ)21x = 2{\left( {\tan \theta } \right)^2} + {\left( {\sin \theta } \right)^2} - 1

Now, let us substitute the value of θ=450\theta = {45^0} in the above expression in order to find the value of x.

x=2(tanθ)2+(sinθ)21 x=2(tan450)2+(sin450)21 (3)  \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
tan450=1\tan {45^0} = 1 and sin450=12\sin {45^0} = \dfrac{1}{{\sqrt 2 }}

Putting these values in equation (3), we get
x=2(1)2+(12)21=2+121=1+12=2+12=32\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(tanθ)2+(sinθ)21=322{\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 θ\theta because tanθ=1\tan \theta = 1 gives various values of θ\theta as θ=450,2250,4050\theta = {45^0},{225^0},{405^0}, etc but in the problem it is given that θ\theta is an acute angle so we will consider only that value of θ\theta which measures less than 900{90^0}. That’s why the only possible result of tanθ=1\tan \theta = 1 is θ=450\theta = {45^0}.