Question

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

Answer 1

Hint: We have a trigonometric expression as: $\sum _{m=1}^6\text{cosec}(\theta +{\displaystyle \frac{(m-1)\pi }{4}})\text{cosec}(\theta +{\displaystyle \frac{m\pi }{4}})=4\sqrt{2}$
We can write $\sqrt{2}=\text{cosec}{\displaystyle \frac{\pi }{4}}$ . As the expression contains $\text{cosec}\theta$ , try to convert the expression in terms of $\sin \theta$ . Then, we can write $\sin {\displaystyle \frac{\pi }{4}}$ as $\sin [(\theta +{\displaystyle \frac{m\pi }{4}})-(\theta +{\displaystyle \frac{(m-1)\pi }{4}})]$
Later on, by using the identity: $\sin (A-B)=\sin A\cos B-\cos A\sin B$ , split the term $\sin [(\theta +{\displaystyle \frac{m\pi }{4}})-(\theta +{\displaystyle \frac{(m-1)\pi }{4}})]$. 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 $\theta$

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

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.