Question

For $0 < x < \dfrac{\pi }{2}$, the solution of $\sum\limits_{m = 1}^6 {\cos ec\left( {\theta + \dfrac{{\left( {m - 1} \right)\pi }}{4}} \right)} \cos ec\left( {\theta + \dfrac{{m\pi }}{4}} \right) = 4\sqrt 2$is
A. $\dfrac{\pi }{4}$
B .$\dfrac{\pi }{{12}}$
C. $\dfrac{\pi }{6}$
D. $\dfrac{{5\pi }}{{12}}$

Answer 1

In trigonometry sin, cos and tan values are the primary functions we consider while solving trigonometric problems. These trigonometry values are used to measure the angles and sides of a right-angle triangle. Apart from sine, cosine and tangent values, other values are cotangent, secant and cosecant. Hence, to solve the given equation we need to apply trigonometric functions and identity formulas to evaluate the terms, and get the value.

Complete step by step answer:
Given,
$\sum\limits_{m = 1}^6 {\cos ec\left( {\theta + \dfrac{{\left( {m - 1} \right)\pi }}{4}} \right)} \cos ec\left( {\theta + \dfrac{{m\pi }}{4}} \right) = 4\sqrt 2$
The given equation is rewritten as:
We, know that $\dfrac{1}{{\sin \theta }} = \cos ec\theta$, hence the given equation is rewritten as:
$\Rightarrow \sum\limits_{m = 1}^6 {\dfrac{1}{{\sin \left( {\theta + \dfrac{{\left( {m - 1} \right)\pi }}{4}} \right)\sin \left( {\theta + \dfrac{{m\pi }}{4}} \right)}}} = 4\sqrt 2$
Multiplying and dividing the equation with $\sin \dfrac{\pi }{4}$ terms of the denominator, we get:
\Rightarrow \sum\limits_{m = 1}^6 {\dfrac{{\sin \left[ {\theta + \dfrac{{m\pi }}{4} - \left( {\theta + \dfrac{{\left( {m - 1} \right)\pi }}{4}} \right)} \right]}}{{\sin \dfrac{\pi }{4}\left\\{ {\sin \left( {\theta + \dfrac{{\left( {m - 1} \right)\pi }}{4}} \right)\sin \left( {\theta + \dfrac{{m\pi }}{4}} \right)} \right\\}}}} = 4\sqrt 2
Simplifying the terms with respect to trigonometric identities function we get:
$\Rightarrow \sum\limits_{m = 1}^6 {\dfrac{{\cot \left( {\theta + \dfrac{{\left( {m - 1} \right)\pi }}{4}} \right) - \cot \left( {\theta + \dfrac{{m\pi }}{4}} \right)}}{{\dfrac{1}{{\sqrt 2 }}}}} = 4\sqrt 2$
We have the common terms in numerator and denominator i.e., $\sqrt 2$, hence we get:
$\Rightarrow \sum\limits_{m = 1}^6 {\left[ {\cot \left( {\theta + \dfrac{{\left( {m - 1} \right)\pi }}{4}} \right) - \cot \left( {\theta + \dfrac{{m\pi }}{4}} \right)} \right]} = 4$
Now, apply summation to the terms as:
$\Rightarrow \cot \left( \theta \right) - \cot \left( {\theta + \dfrac{\pi }{4}} \right) + \cot \left( {\theta + \dfrac{\pi }{4}} \right) - \cot \left( {\theta + \dfrac{{2\pi }}{4}} \right) + ..... + \cot \left( {\theta + \dfrac{{5\pi }}{4}} \right) - \cot \left( {\theta + \dfrac{{6\pi }}{4}} \right) = 4$
Simplifying the terms, we get:
$\Rightarrow \cot \left( \theta \right) - \cot \left( {\theta + \dfrac{{3\pi }}{2}} \right) = 4$
As $\theta$lies in fourth quadrant we have:
$\Rightarrow \cot \theta + \tan \theta = 4$
We know that, $\cot \theta = \dfrac{1}{{\tan \theta }}$, hence applying we get:
$\Rightarrow \dfrac{1}{{\tan }} + \tan \theta = 4$
We get$\Rightarrow {\tan ^2}\theta - 4\tan \theta + 1 = 0$
As, it is of the form $a{x^2} + bx + c$, we have the formula as $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, hence we get:
$\Rightarrow \tan \theta = \dfrac{{4 \pm \sqrt {16 - 4} }}{2}$
$\Rightarrow \tan \theta = \dfrac{{4 \pm \sqrt {12} }}{2} = \dfrac{{4 \pm 2\sqrt 3 }}{2}$
$\Rightarrow \tan \theta = 2 \pm \sqrt 3$
or $\tan \theta = 2 - \sqrt 3$ and $\tan \theta = 2 + \sqrt 3$
Hence, to find the angle $\theta$, we have:
$\Rightarrow \theta = {\tan ^{ - 1}}\left( {2 \pm \sqrt 3 } \right)$
As, we have $2 \pm \sqrt 3$, hence we get:
$\Rightarrow \theta = \dfrac{\pi }{{12}}$and $\theta = \dfrac{{5\pi }}{{12}}$
So, the correct answer is Option B,D

Note: We must know the Trigonometric function chart which helps us to find the values of the function very easily. The key point to find the values of any trigonometric function is to note the chart of all functions as shown and calculate all the terms asked. And here are some of the formulas to be noted.
$\dfrac{1}{{\sin \theta }} = \cos ec\theta$
$\cot \theta = \dfrac{1}{{\tan \theta }}$
$\sin \theta = \dfrac{{\tan \theta }}{{\sec \theta }}$
$\cos \theta = \dfrac{{\sin \theta }}{{\tan \theta }}$
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$