Question

For $0 < \theta < \dfrac{\pi }{2}$ 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/are:

$$A.{\text{ }}\dfrac{\pi }{4} \\\ B.{\text{ }}\dfrac{\pi }{6} \\\ C.{\text{ }}\dfrac{\pi }{{12}} \\\ D.{\text{ }}\dfrac{{5\pi }}{{12}} \\\$$

Answer 1

In order to solve the given summation term for the given set of values, first we will convert the cosec term into sine term, then we will expand the summation by putting the value from 1 to 6 in the place of m and further we will simplify the terms by the use of trigonometric formulas and also use the concept of quadratic equation to get the final answer.

Complete step-by-step answer :
Given that:
$\sum\limits_{m = 1}^6 {\cos ec\left( {\theta + \dfrac{{\left( {m - 1} \right)\pi }}{4}} \right)\cos ec\left( {\theta + \dfrac{{m\pi }}{4}} \right)} = 4\sqrt 2$
In order to simplify the term first let us convert the cosec term to sine term, so we get:
$\Rightarrow \sum\limits_{m = 1}^6 {\dfrac{1}{{\sin \left( {\theta + \dfrac{{\left( {m - 1} \right)\pi }}{4}} \right)\sin \left( {\theta + \dfrac{{m\pi }}{4}} \right)}}} = 4\sqrt 2$
Now let us expand the terms by substituting the value 1 to 6 in the place of m.
$\Rightarrow \left[ {\dfrac{1}{{\sin \left( \theta \right)\sin \left( {\theta + \dfrac{\pi }{4}} \right)}} + \dfrac{1}{{\sin \left( {\theta + \dfrac{\pi }{4}} \right)\sin \left( {\theta + \dfrac{\pi }{2}} \right)}} + ....... + \dfrac{1}{{\sin \left( {\theta + \dfrac{{5\pi }}{4}} \right)\sin \left( {\theta + \dfrac{{3\pi }}{2}} \right)}}} \right] = 4\sqrt 2$
Now let us manipulate the numerator of each fraction as a sum of denominator terms.

$$\Rightarrow \dfrac{1}{{\sin \dfrac{\pi }{4}}}\left[ {\dfrac{{\sin \dfrac{\pi }{4}}}{{\sin \left( \theta \right)\sin \left( {\theta + \dfrac{\pi }{4}} \right)}} + \dfrac{{\sin \dfrac{\pi }{4}}}{{\sin \left( {\theta + \dfrac{\pi }{4}} \right)\sin \left( {\theta + \dfrac{\pi }{2}} \right)}} + ....... + \dfrac{{\sin \dfrac{\pi }{4}}}{{\sin \left( {\theta + \dfrac{{5\pi }}{4}} \right)\sin \left( {\theta + \dfrac{{3\pi }}{2}} \right)}}} \right] = 4\sqrt 2 \\\ \Rightarrow \sqrt 2 \left[ {\dfrac{{\sin \left( {\theta + \dfrac{\pi }{4} - \theta } \right)}}{{\sin \left( \theta \right)\sin \left( {\theta + \dfrac{\pi }{4}} \right)}} + \dfrac{{\sin \left( {\theta + \dfrac{\pi }{2} - \left( {\theta + \dfrac{\pi }{4}} \right)} \right)}}{{\sin \left( {\theta + \dfrac{\pi }{4}} \right)\sin \left( {\theta + \dfrac{\pi }{2}} \right)}} + ....... + \dfrac{{\sin \left( {\theta + \dfrac{{3\pi }}{2} - \left( {\theta + \dfrac{{5\pi }}{4}} \right)} \right)}}{{\sin \left( {\theta + \dfrac{{5\pi }}{4}} \right)\sin \left( {\theta + \dfrac{{3\pi }}{2}} \right)}}} \right] = 4\sqrt 2 \\\$$

As we know the trigonometric identity for the sum of angle of sine is given as:
$\sin \left( {a - b} \right) = \sin a\cos b - \cos a\sin b$
Now let use the given identity for sum of sine of angles in the place of numerator in each of the terms given above:
$\Rightarrow \left[ {\dfrac{{\sin \left( {\theta + \dfrac{\pi }{4}} \right)\cos \theta - \cos \left( {\theta + \dfrac{\pi }{4}} \right)\sin \theta }}{{\sin \left( \theta \right)\sin \left( {\theta + \dfrac{\pi }{4}} \right)}} + \dfrac{{\sin \left( {\theta + \dfrac{\pi }{2}} \right)\cos \left( {\theta + \dfrac{\pi }{4}} \right) - \cos \left( {\theta + \dfrac{\pi }{2}} \right)\sin \left( {\theta + \dfrac{\pi }{4}} \right)}}{{\sin \left( {\theta + \dfrac{\pi }{4}} \right)\sin \left( {\theta + \dfrac{\pi }{2}} \right)}} + .......} \right] = 4$Let us now simplify the above term to get terms as cotangent and cancel some terms

\Rightarrow \left[ {\dfrac{{\sin \left( {\theta + \dfrac{\pi }{4}} \right)\cos \theta }}{{\sin \left( \theta \right)\sin \left( {\theta + \dfrac{\pi }{4}} \right)}} - \dfrac{{\cos \left( {\theta + \dfrac{\pi }{4}} \right)\sin \theta }}{{\sin \left( \theta \right)\sin \left( {\theta + \dfrac{\pi }{4}} \right)}} + \dfrac{{\sin \left( {\theta + \dfrac{\pi }{2}} \right)\cos \left( {\theta + \dfrac{\pi }{4}} \right)}}{{\sin \left( {\theta + \dfrac{\pi }{4}} \right)\sin \left( {\theta + \dfrac{\pi }{2}} \right)}} - \dfrac{{\cos \left( {\theta + \dfrac{\pi }{2}} \right)\sin \left( {\theta + \dfrac{\pi }{4}} \right)}}{{\sin \left( {\theta + \dfrac{\pi }{4}} \right)\sin \left( {\theta + \dfrac{\pi }{2}} \right)}} + .......} \right] = 4 \\\ \Rightarrow \left[ {\cot \theta - \cot \left( {\theta + \dfrac{\pi }{4}} \right) + \cot \left( {\theta + \dfrac{\pi }{4}} \right) - \cot \left( {\theta + \dfrac{\pi }{2}} \right) + ....... + \cot \left( {\theta + \dfrac{{5\pi }}{4}} \right) - \cot \left( {\theta + \dfrac{{3\pi }}{2}} \right)} \right] = 4 \\\ \Rightarrow \left[ {\cot \theta - \cot \left( {\theta + \dfrac{{3\pi }}{2}} \right)} \right] = 4 \\\ $$As we know the formula for the conversion for cotangent is given as: $$\cot \left( {\theta + \dfrac{{3\pi }}{2}} \right) = - \tan \theta $$ So let us use the same in the above term to get the equation as: $$ \Rightarrow \cot \theta + \tan \theta = 4$$ Let us now convert cotangent into tangent to proceed further to find the value of angle:

\Rightarrow \dfrac{1}{{\tan \theta }} + \tan \theta = 4 \\
\Rightarrow {\tan ^2}\theta - 4\tan \theta + 1 = 0 \\

Let us now consider the given equation as the equation in variable tangent theta and let solve for tangent theta. $$ \Rightarrow \tan \theta = \dfrac{{4 \pm \sqrt {{4^2} - 4 \times 1 \times 1} }}{{2 \times 1}}\left[ {\because x = \dfrac{{ - b \pm \sqrt {{b^2} - 4 \times a \times c} }}{{2 \times a}}} \right]$$ Let us simplify the term to get the value of theta

\Rightarrow \tan \theta = \dfrac{{4 \pm \sqrt {16 - 4} }}{2} \\
\Rightarrow \tan \theta = \dfrac{{4 \pm \sqrt {12} }}{2} \\
\Rightarrow \tan \theta = \dfrac{{4 \pm 2\sqrt 3 }}{2} \\
\Rightarrow \tan \theta = 2 \pm \sqrt 3 \\

As we have the value of tangent of theta and also the range of theta which is given as: $$0 < \theta < \dfrac{\pi }{2}$$ So, the values are: $ \Rightarrow \theta = \dfrac{\pi }{{12}}\& \dfrac{{5\pi }}{{12}}$ Hence the value of $\theta = \dfrac{\pi }{{12}},\dfrac{{5\pi }}{{12}}$ **So, the correct options are C and D.** **Note** : In order to solve such types of problems students must remember the basic method of expansion of series and should also remember the formulas for the conversion of trigonometric identities from one form to another. Students must take extra care of the range of angles provided.