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# For $0<\theta <\dfrac{\pi }{2}$ , the solution(s) of $\sum\limits_{m=1}^{6}{\text{cosec}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)}\text{cosec}\left( \theta +\dfrac{m\pi }{4} \right)=4\sqrt{2}$ is (are):a)$\dfrac{\pi }{4}$b)$\dfrac{\pi }{6}$ c)$\dfrac{\pi }{12}$ d) $\dfrac{5\pi }{12}$

Last updated date: 19th Jun 2024
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Hint: We have a trigonometric expression as: $\sum\limits_{m=1}^{6}{\text{cosec}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)}\text{cosec}\left( \theta +\dfrac{m\pi }{4} \right)=4\sqrt{2}$
We can write $\sqrt{2}=\text{cosec}\dfrac{\pi }{4}$ . As the expression contains $\text{cosec}\theta$ , try to convert the expression in terms of $\sin \theta$ . Then, we can write $\sin \dfrac{\pi }{4}$ as $\sin \left[ \left( \theta +\dfrac{m\pi }{4} \right)-\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right) \right]$
Later on, by using the identity: $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ , split the term $\sin \left[ \left( \theta +\dfrac{m\pi }{4} \right)-\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right) \right]$. Now, simplify the whole expression by cancelling the terms to get an equation in terms of $\cot \theta$ . Now, expand the summation given by putting values of m and cancel out to the terms to get a simplified equation. Now, using various trigonometric identities, find the value of $\theta$

We have:
$\sum\limits_{m=1}^{6}{\text{cosec}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)}\text{cosec}\left( \theta +\dfrac{m\pi }{4} \right)=4\sqrt{2}......(1)$
As we know that: $\text{cosec}\dfrac{\pi }{4}=\sqrt{2}$
So, we can write equation (1) as:
$\sum\limits_{m=1}^{6}{\text{cosec}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)}\text{cosec}\left( \theta +\dfrac{m\pi }{4} \right)=4\text{cosec}\dfrac{\pi }{4}......(2)$
As we know that: $\text{cosec}\theta =\dfrac{1}{\sin \theta }$ , so we can write equation (2) as:
\begin{align} & \sum\limits_{m=1}^{6}{\dfrac{1}{\text{sin}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)\text{sin}\left( \theta +\dfrac{m\pi }{4} \right)}}=\dfrac{4}{\text{sin}\dfrac{\pi }{4}} \\ & \sum\limits_{m=1}^{6}{\dfrac{\text{sin}\dfrac{\pi }{4}}{\text{sin}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)\text{sin}\left( \theta +\dfrac{m\pi }{4} \right)}}=4......(3) \\ \end{align}
Now, we can write: $\sin \dfrac{\pi }{4}=\sin \left[ \left( \theta +\dfrac{m\pi }{4} \right)-\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right) \right]$ in equation (3), we get:
$\sum\limits_{m=1}^{6}{\dfrac{\sin \left[ \left( \theta +\dfrac{m\pi }{4} \right)-\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right) \right]}{\text{sin}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)\text{sin}\left( \theta +\dfrac{m\pi }{4} \right)}}=4......(4)$
Now, by applying identity: $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$, we can write equation (4) as:
$\sum\limits_{m=1}^{6}{\dfrac{\left[ \sin \left( \theta +\dfrac{m\pi }{4} \right)\cos \left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)-\cos \left( \theta +\dfrac{m\pi }{4} \right)\sin \left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right) \right]}{\text{sin}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)\text{sin}\left( \theta +\dfrac{m\pi }{4} \right)}}=4......(5)$
Now, by expanding the equation (5), we get:
\begin{align} & \sum\limits_{m=1}^{6}{\dfrac{\sin \left( \theta +\dfrac{m\pi }{4} \right)\cos \left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)}{\text{sin}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)\text{sin}\left( \theta +\dfrac{m\pi }{4} \right)}}-\dfrac{\cos \left( \theta +\dfrac{m\pi }{4} \right)\sin \left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)}{\text{sin}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)\text{sin}\left( \theta +\dfrac{m\pi }{4} \right)}=4 \\ & \sum\limits_{m=1}^{6}{\dfrac{\cos \left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)}{\text{sin}\left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)}}-\dfrac{\cos \left( \theta +\dfrac{m\pi }{4} \right)}{\text{sin}\left( \theta +\dfrac{m\pi }{4} \right)}=4......(6) \\ \end{align}
Since $\dfrac{\cos \theta }{\sin \theta }=\cot \theta$ , we can write equation (6) as:
$\sum\limits_{m=1}^{6}{\cot \left( \theta +\dfrac{\left( m-1 \right)\pi }{4} \right)}-\cot \left( \theta +\dfrac{m\pi }{4} \right)=4......(7)$
Now, expand the summation by putting the values of m, we get:
\begin{align} & \Rightarrow \left[ \cot \left( \theta +\dfrac{\left( 1-1 \right)\pi }{4} \right)-\cot \left( \theta +\dfrac{\pi }{4} \right) \right]+\left[ \cot \left( \theta +\dfrac{\left( 2-1 \right)\pi }{4} \right)-\cot \left( \theta +\dfrac{2\pi }{4} \right) \right] \\ & \text{ }+.....+\left[ \cot \left( \theta +\dfrac{\left( 6-1 \right)\pi }{4} \right)-\cot \left( \theta +\dfrac{6\pi }{4} \right) \right]=4 \\ & \Rightarrow \cot \theta -\cot \left( \theta +\dfrac{\pi }{4} \right)+\cot \left( \theta +\dfrac{\pi }{4} \right)-\cot \left( \theta +\dfrac{2\pi }{4} \right) \\ & \text{ }+.....+\cot \left( \theta +\dfrac{5\pi }{4} \right)-\cot \left( \theta +\dfrac{6\pi }{4} \right)=4 \\ & \Rightarrow \cot \theta -\cot \left( \theta +\dfrac{6\pi }{4} \right)=4 \\ & \Rightarrow \cot \theta -\cot \left( \theta +\dfrac{3\pi }{2} \right)=4......(8) \\ \end{align}
As we know that: $\cot \left( \dfrac{3\pi }{2}+\theta \right)=-\tan \theta$
So, we can write equation (8) as:
$\Rightarrow \cot \theta +\tan \theta =4......(9)$
Now, write $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ and $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ in equation (9), we get:
\begin{align} & \Rightarrow \cot \theta +\tan \theta =4 \\ & \Rightarrow \dfrac{\cos \theta }{\sin \theta }+\dfrac{\sin \theta }{\cos \theta }=4 \\ & \Rightarrow {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =4\sin \theta \cos \theta ......(10) \\ \end{align}
As we know that: ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ and $2\sin \theta \cos \theta =\sin 2\theta$, so we can write equation (10) as:
$\Rightarrow 1=2\sin 2\theta ......(11)$
Now, solving for $\theta$, we can write equation (11) as:
\begin{align} & \Rightarrow \sin 2\theta =\dfrac{1}{2} \\ & \Rightarrow \sin 2\theta =\sin \dfrac{\pi }{6}\text{ or }\sin \dfrac{5\pi }{6} \\ & \Rightarrow 2\theta =\dfrac{\pi }{6}\text{ or }\dfrac{5\pi }{6} \\ & \Rightarrow \theta =\dfrac{\pi }{12}\text{ or }\dfrac{5\pi }{12} \\ \end{align}

So, the correct answer is “Option C and D”.

Note: For a given trigonometric expression, it is always easier to convert the expression in terms of sine and cosine. Also, if a summation expression is given, always try to expand the summation by putting the values of the variable and cancel out the terms if possible.