Question

The value of ${\sec^{ - 1}}\left( {\dfrac{1}{4}\sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)} } \right)$ in the interval $\left[ { - \dfrac{\pi }{4},\dfrac{{3\pi }}{4}} \right]$ equals

Answer 1

Hint: Here we have to find the value of the given function. For that, we first simplify the given function and then apply formula $\sec \left( {\dfrac{\pi }{2}} \right) = - {\text{cosec}}\theta $ and multiply 2 in the numerator and denominator and convert ${\text{cosec}}\theta$ to $\dfrac{1}{{\sin \theta }}$ in the denominator and simplify it to calculate the desired result.

Formula Used:
We have been using the following formulas:
1. $\sec \left( {\dfrac{\pi }{2}} \right) = - {\text{cosec}}\theta $
2. $2\sin \theta \cos \theta = \sin 2\theta $
3. ${\text{cosec}}\theta = \dfrac{1}{{\sin \theta }}$
4. $\sec \theta = \dfrac{1}{{\cos \theta }}$

Complete step by step solution:
Given that ${\sec^{ - 1}}\left( {\dfrac{1}{4}\sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)} } \right)$
By simplifying the given function, we get
${\sec^{ - 1}}\left( {\dfrac{1}{4}\sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)} } \right) = {\sec^{ - 1}}\left( {\dfrac{1}{4}\sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi + \pi }}{2}} \right)} } \right) \\ \Rightarrow {\sec^{ - 1}}\left( {\dfrac{1}{4}\sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2} + \dfrac{\pi }{2}} \right)} } \right) \\ $
W.K.T $\sec \left( {\dfrac{\pi }{2}} \right) = - {\text{cosec}}\theta $
So, by applying this formula, we get
${\sec^{ - 1}}\left( {\dfrac{1}{4}\sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)} } \right) = {\sec^{ - 1}}\left( { - \dfrac{1}{4}\sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right){\text{cosec}}\left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)} } \right)$
Here, $\theta = \dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}$
We know that $\sec \theta = \dfrac{1}{{\cos \theta }}$ and ${\text{cosec}}\theta = \dfrac{1}{{\sin \theta }}$
So, our function becomes
${\sec^{ - 1}}\left( {\dfrac{1}{4}\sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)} } \right) = {\sec^{ - 1}}\left( { - \dfrac{1}{4}\sum\limits_{k = 0}^{10} {\dfrac{1}{{\cos \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sin \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)}}} } \right)$
Multiply 2 in both numerator and denominator, we get
${\sec^{ - 1}}\left( {\dfrac{1}{4}\sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)} } \right) = {\sec^{ - 1}}\left( { - \dfrac{1}{4}\sum\limits_{k = 0}^{10} {\dfrac{{1 \times 2}}{{2\cos \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sin \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)}}} } \right)$
We know that $2\sin \theta \cos \theta = \sin 2\theta $
Applying this formula in the denominator of the function, we get
${\sec^{ - 1}}\left( {\dfrac{1}{4}\sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)} } \right) = {\sec^{ - 1}}\left( { - \dfrac{1}{2}\sum\limits_{k = 0}^{10} {\dfrac{1}{{\sin 2\left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)}}} } \right) \\ \Rightarrow {\sec^{ - 1}}\left( { - \dfrac{1}{2}\sum\limits_{k = 0}^{10} {\dfrac{1}{{\sin \left( {\dfrac{{7\pi }}{6} + k\pi } \right)}}} } \right) \\ \Rightarrow {\sec^{ - 1}}\left( { - \dfrac{1}{2}\dfrac{1}{{\sin \left( {k\pi + \pi + \dfrac{\pi }{6}} \right)}}} \right) \\ \Rightarrow {\sec^{ - 1}}\left( { - \dfrac{1}{2}\dfrac{1}{{\sin \left( {\left( {k + 1} \right)\pi + \dfrac{\pi }{6}} \right)}}} \right) \\ $
Further simplifying, we get
${\sec^{ - 1}}\left( {\dfrac{1}{4}\sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)} } \right) = {\sec^{ - 1}}\left( { - \dfrac{1}{2}\sum\limits_{k = 0}^{10} {\dfrac{1}{{{{\left( { - 1} \right)}^{k + 1}}\sin \left( {\dfrac{\pi }{6}} \right)}}} } \right) \\ \Rightarrow {\sec^{ - 1}}\left( { - \dfrac{1}{2}\sum\limits_{k = 0}^{10} {\dfrac{1}{{{{\left( { - 1} \right)}^{k + 1}} \times \dfrac{1}{2}}}} } \right) \\ \Rightarrow {\sec^{ - 1}}\left( { - \dfrac{1}{2}\sum\limits_{k = 0}^{10} {\dfrac{{1 \times 2}}{{{{\left( { - 1} \right)}^{k + 1}} \times 1}}} } \right) \\ \Rightarrow {\sec^{ - 1}}\left( { - 1\sum\limits_{k = 0}^{10} {\dfrac{1}{{{{\left( { - 1} \right)}^{k + 1}}}}} } \right) \\ $
Furthermore simplifying, we get
$ {\sec^{ - 1}}\left( {\dfrac{1}{4}\sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)} } \right) = {\sec^{ - 1}}\left( 1 \right) \\ \Rightarrow {\sec^{ - 1}}\left( {\sec {0^ \circ }} \right) $

Hence, the value of ${\sec^{ - 1}}\left( {\dfrac{1}{4}\sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)} } \right)$ is equal to $0$

Note: Alternative method to solve the above problem:
Given that ${\sec^{ - 1}}\left( {\dfrac{1}{4}\sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)} } \right)...\left( 1 \right)$
First solve $\sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)} ...\left( 2 \right)$
W.K.T $\sec \theta = \dfrac{1}{{\cos \theta }}$
Applying the above formula in equation (2), we get
$\sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)} = \sum\limits_{k = 0}^{10} {\dfrac{1}{{\cos \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\cos \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)}}} \\ \Rightarrow \sum\limits_{k = 1}^{10} {\dfrac{{\sin \left[ {\left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right) - \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)} \right]}}{{\cos \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\cos \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)}}} \\ $
$\left[ \because \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right) - \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right) = \dfrac{\pi }{2} \\ \sin \left( {\dfrac{\pi }{2}} \right) = 1 \\ \right]$

We know that $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$
By applying above formula, we get
$ \sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)} = \sum\limits_{k = 0}^{10} {\dfrac{{\sin \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)\cos \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right) - \cos \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)\sin \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)}}{{\cos \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\cos \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)}}} \\ \Rightarrow \sum\limits_{k = 0}^{10} {\dfrac{{\sin \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)\cos \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)}}{{\cos \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\cos \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)}}} - \dfrac{{\cos \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)\sin \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)}}{{\cos \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\cos \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)}} \\ \Rightarrow \sum\limits_{k = 0}^{10} {\dfrac{{\sin \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)}}{{\cos \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)}}} - \dfrac{{\sin \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)}}{{\cos \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)}}$

We know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
By applying above formula, we get
$\sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)} = \sum\limits_{k = 0}^{10} {\left[ {\tan \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right) - \tan \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)} \right]} \\ \Rightarrow \tan \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2} + \dfrac{\pi }{2}} \right) - \tan \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right) \\ \Rightarrow \tan \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right) + \tan \left( {\dfrac{\pi }{2}} \right) - \tan \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right) \\ \Rightarrow \tan \left( {\dfrac{\pi }{2}} \right)$

Further solving, we get
$\sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)} = 0$
Substitute this value in equation (1), we get
${\sec^{ - 1}}\left( {\dfrac{1}{4}\sum\limits_{k = 0}^{10} {\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{k\pi }}{2}} \right)\sec \left( {\dfrac{{7\pi }}{{12}} + \dfrac{{\left( {k + 1} \right)\pi }}{2}} \right)} } \right) = {\sec^{ - 1}}\left( {\dfrac{1}{4} \times 0} \right) \\ \Rightarrow {\sec^{ - 1}}\left( 0 \right) = 0$