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

Last updated date: 13th Jun 2024
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Hint: We use the given equality between sine and cosine of an angle. Divide both sides of the equality with such a trigonometric function so we can make out the value of angle using tangent of the angle. Find the angle and substitute it in the equation whose value we have to find. Check whether the angle is acute or not. Use a table of trigonometric values to substitute values in the equation.
* Acute angle: Any angle having measure less than ${90^ \circ }$ is called an acute angle.
* $\tan x = \dfrac{{\sin x}}{{\cos x}}$

Complete step-by-step solution:
We are given $\sin \theta = \cos \theta$....................… (1)
We divide both sides of equation (1) by $\cos \theta$
$\Rightarrow \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{\cos \theta }}{{\cos \theta }}$
Cancel same terms in RHS of the equation and use $\tan x = \dfrac{{\sin x}}{{\cos x}}$in LHS of the equation
$\Rightarrow \tan \theta = 1$........................… (2)
Now we know from the table of trigonometric functions that$\tan {45^ \circ } = 1$. Substitute the value of 1 in equation (2)
$\Rightarrow \tan \theta = \tan {45^ \circ }$
Take inverse tangent function on both sides of the equation
$\Rightarrow {\tan ^{ - 1}}\left( {\tan \theta } \right) = {\tan ^{ - 1}}\left( {\tan {{45}^ \circ }} \right)$
Since we know ${f^{ - 1}}(f(x)) = x$
$\Rightarrow \theta = {45^ \circ }$.........................… (3)
Since, $\theta = {45^ \circ } < {90^ \circ }$, so the angle is an acute angle.
Now substitute the value from equation (3) in the equation $2{\tan ^2}\theta + {\sin ^2}\theta - 1$
$\Rightarrow 2{\tan ^2}\theta + {\sin ^2}\theta - 1 = 2{\tan ^2}({45^ \circ }) + {\sin ^2}({45^ \circ }) - 1$
$\Rightarrow 2{\tan ^2}\theta + {\sin ^2}\theta - 1 = 2{\left[ {\tan ({{45}^ \circ })} \right]^2} + {\left[ {\sin ({{45}^ \circ })} \right]^2} - 1$.................… (4)
Substitute the value of $\tan {45^ \circ } = 1,\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$ in equation (4)
$\Rightarrow 2{\tan ^2}\theta + {\sin ^2}\theta - 1 = 2{(1)^2} + {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2} - 1$
$\Rightarrow 2{\tan ^2}\theta + {\sin ^2}\theta - 1 = 2 + \dfrac{1}{2} - 1$
Take LCM in RHS of the equation
$\Rightarrow 2{\tan ^2}\theta + {\sin ^2}\theta - 1 = \dfrac{{4 + 1 - 2}}{2}$
$\Rightarrow 2{\tan ^2}\theta + {\sin ^2}\theta - 1 = \dfrac{3}{2}$

$\therefore$The value of $2{\tan ^2}\theta + {\sin ^2}\theta - 1$ is $\dfrac{3}{2}$

Note: * The table that tells us some basic values of trigonometric functions at common angles is given as
 Angles (in degrees) ${0^ \circ }$ ${30^ \circ }$ ${45^ \circ }$ ${60^ \circ }$ ${90^ \circ }$ sin 0 $\dfrac{1}{2}$ $\dfrac{1}{{\sqrt 2 }}$ $\dfrac{{\sqrt 3 }}{2}$ $1$ cos 1 $\dfrac{{\sqrt 3 }}{2}$ $\dfrac{1}{{\sqrt 2 }}$ $\dfrac{1}{2}$ 0 tan 0 $\dfrac{1}{{\sqrt 3 }}$ 1 $\sqrt 3$ Not defined cosec Not defined 2 $\sqrt 2$ $\dfrac{2}{{\sqrt 3 }}$ 1 sec 1 $\dfrac{2}{{\sqrt 3 }}$ $\sqrt 2$ 2 Not defined cot Not defined $\sqrt 3$ 1 $\dfrac{1}{{\sqrt 3 }}$ 0

* Inverse of any function when applied on the same function cancels out the function and the inverse, i.e. ${f^{ - 1}}(f(x)) = x$.