Question

If $\dfrac{\pi }{2} < \theta < \dfrac{{3\pi }}{2}$, then the value of $\sqrt {4{{\cos }^4}\theta + {{\sin }^2}2\theta } + 4\cot \theta {\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$ is
(A) $2\sin \theta$
(B) $- 2\sin \theta$
(C) $2\cot \theta$
(D) $- 2\cot \theta$

Answer 1

In this question, we have to evaluate the value of a given particular in the given region.
We need to first put the trigonometric formulas to bring it in a shorter form so that we can get a solution, then getting the square root of the square of the cosine function in given region we will get the negative of cosine then putting the algebraic and trigonometric formulas we will get the solution.

Formula used: Here we have used the algebraic formula,
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$.
The trigonometric formulas we have used,
$\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
$\sin 2\theta = 2\sin \theta \cos \theta$
$\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$

Complete step-by-step answer:
We need to find out the values of $\sqrt {4{{\cos }^4}\theta + {{\sin }^2}2\theta } + 4\cot \theta {\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$.
Now,$\sqrt {4{{\cos }^4}\theta + {{\sin }^2}2\theta } + 4\cot \theta {\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$
Using the trigonometric formula,$\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$,we get,
$\Rightarrow \sqrt {4{{\cos }^4}\theta + 4{{\sin }^2}\theta {{\cos }^2}\theta } + 4\cot \theta {\left( {\cos \dfrac{\pi }{4}\cos \dfrac{\theta }{2} + \sin \dfrac{\pi }{4}\sin \dfrac{\theta }{2}} \right)^2}$
Putting the value of $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ , $\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$, we get,
$\Rightarrow \sqrt {4{{\cos }^4}\theta + 4{{\sin }^2}\theta {{\cos }^2}\theta } + 4\cot \theta {\left( {\dfrac{1}{{\sqrt 2 }}\cos \dfrac{\theta }{2} + \dfrac{1}{{\sqrt 2 }}\sin \dfrac{\theta }{2}} \right)^2}$
Since, $\dfrac{\pi }{2} < \theta < \dfrac{{3\pi }}{2}$, $\theta$ is in second or third quadrant so the value of $\cos \theta$ is negative.
Thus, $\sqrt {{{\cos }^2}\theta } = - \cos \theta$, we get,
$\Rightarrow - 2\cos \theta \sqrt {{{\cos }^2}\theta + {{\sin }^2}\theta } + 4\cot \theta \dfrac{1}{2}{\left( {\cos \dfrac{\theta }{2} + \sin \dfrac{\theta }{2}} \right)^2}$
Using the algebraic formula,${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$
$\Rightarrow - 2\cos \theta + 2\cot \theta \left( {{{\cos }^2}\dfrac{\theta }{2} + {{\sin }^2}\dfrac{\theta }{2} + 2\cos \dfrac{\theta }{2}\sin \dfrac{\theta }{2}} \right)$
Using the trigonometric formula, $\sin 2\theta = 2\sin \theta \cos \theta$
$\Rightarrow - 2\cos \theta + 2\cot \theta \left( {1 + \sin \theta } \right)$
Using the formula, $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
$\Rightarrow - 2\cos \theta + 2\cot \theta + 2\dfrac{{\cos \theta }}{{\sin \theta }}\sin \theta$
Cancelling the term $\sin \theta$ in numerator and denominator we get,
$\Rightarrow - 2\cos \theta + 2\cot \theta + 2\cos \theta$
Simplifying we get,
$\Rightarrow 2\cot \theta$
Hence we get, $\sqrt {4{{\cos }^4}\theta + {{\sin }^2}2\theta } + 4\cot \theta {\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right) = 2\cot \theta$ in $\dfrac{\pi }{2} < \theta < \dfrac{{3\pi }}{2}$.

$\therefore$Thus (C) is the correct option.

Note:

In the first quadrant $\left( {0{\text{ to }}\dfrac{\pi }{2}} \right)$ all trigonometric functions are positive, in second quadrant $\left( {\dfrac{\pi }{2}{\text{ to }}\pi } \right)$ only sine function is positive, and in third quadrant $\left( {\pi {\text{ to }}\dfrac{{3\pi }}{2}} \right)$ tangent function is positive, in fourth quadrant $\left( {\dfrac{{3\pi }}{2}{\text{ to }}2\pi } \right)$ cosine functions are positive.
Thus if $\dfrac{\pi }{2} < \theta < \dfrac{{3\pi }}{2}$ then $\theta$ lies between second and third quadrant where the value of cosine function is negative. Therefore we get the square root of the cosine square function as negative of the cosine function.
We know that cotangent function is given by dividing cosine function by sine function.