Question

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

Answer 1

To solve this question we will only solve and keep simplifying the brackets by using value of standard trigonometric values and inverse trigonometric values. After the appropriate simplification, we will expand the summation expression and hence, we will end up with ${{\sec }^{-1}}(1)$, which we know is equals to 0.

Complete step-by-step answer:
Now, we have ${{\sec }^{-1}}\left[ \dfrac{1}{4}\sum\limits_{k=0}^{10}{\left( \sec \left( \dfrac{7\pi }{12}+\dfrac{k\pi }{2} \right)\sec \left( \dfrac{7\pi }{12}+(k+1)\dfrac{\pi }{2} \right) \right)} \right]$ and we have to find its value in the interval of $\left[ -\dfrac{\pi }{4},\dfrac{3\pi }{4} \right]$.
So, to get the value we will solve the summation and simplify the brackets.
Now, ${{\sec }^{-1}}\left[ \dfrac{1}{4}\sum\limits_{k=0}^{10}{\left( \sec \left( \dfrac{7\pi }{12}+\dfrac{k\pi }{2} \right)\sec \left( \dfrac{7\pi }{12}+(k+1)\dfrac{\pi }{2} \right) \right)} \right]$
We can write above expression as
${{\sec }^{-1}}\left[ \dfrac{1}{4}\sum\limits_{k=0}^{10}{\left( \sec \left( \dfrac{7\pi }{12}+\dfrac{k\pi }{2} \right)\sec \left( \dfrac{7\pi }{12}+k\dfrac{\pi }{2}+\dfrac{\pi }{2} \right) \right)} \right]$
We know that, $\sec \left( \dfrac{\pi }{2}+\theta \right)=-\cos ec\theta$
So, we can write $\sec \left( \dfrac{7\pi }{12}+k\dfrac{\pi }{2}+\dfrac{\pi }{2} \right)$ as $-cosec\left( \dfrac{7\pi }{12}+k\dfrac{\pi }{2} \right)$
So, we have ${{\sec }^{-1}}\left[ -\dfrac{1}{4}\sum\limits_{k=0}^{10}{\left( \sec \left( \dfrac{7\pi }{12}+\dfrac{k\pi }{2} \right)cosec\left( \dfrac{7\pi }{12}+k\dfrac{\pi }{2} \right) \right)} \right]$
We know that, $\sec x=\dfrac{1}{\cos x}$ and $cosecx=\dfrac{1}{sinx}$
So, we can write $\sec \left( \dfrac{7\pi }{12}+\dfrac{k\pi }{2} \right)$ as $\dfrac{1}{\cos \left( \dfrac{7\pi }{12}+k\dfrac{\pi }{2} \right)}$ and $cosec\left( \dfrac{7\pi }{12}+k\dfrac{\pi }{2} \right)$ as $\dfrac{1}{sin\left( \dfrac{7\pi }{12}+k\dfrac{\pi }{2} \right)}$
So, we have ${{\sec }^{-1}}\left[ -\dfrac{1}{4}\sum\limits_{k=0}^{10}{\left( \dfrac{1}{\cos \left( \dfrac{7\pi }{12}+k\dfrac{\pi }{2} \right)sin\left( \dfrac{7\pi }{12}+k\dfrac{\pi }{2} \right)} \right)} \right]$
Multiplying, both numerator and denominator of summation by 2, we get
${{\sec }^{-1}}\left[ -\dfrac{1}{4}\sum\limits_{k=0}^{10}{\left( \dfrac{2}{2\cos \left( \dfrac{7\pi }{12}+k\dfrac{\pi }{2} \right)sin\left( \dfrac{7\pi }{12}+k\dfrac{\pi }{2} \right)} \right)} \right]$
We know that, $2\sin x\cos x=\sin 2x$ ,
So, we can write denominator $2\cos \left( \dfrac{7\pi }{12}+k\dfrac{\pi }{2} \right)sin\left( \dfrac{7\pi }{12}+k\dfrac{\pi }{2} \right)$ as $sin2\left( \dfrac{7\pi }{12}+k\dfrac{\pi }{2} \right)$
So, we have ${{\sec }^{-1}}\left[ -\dfrac{1}{4}\sum\limits_{k=0}^{10}{\left( \dfrac{2}{sin2\left( \dfrac{7\pi }{12}+k\dfrac{\pi }{2} \right)} \right)} \right]$
On simplifying, we get
${{\sec }^{-1}}\left[ -\dfrac{1}{4}\sum\limits_{k=0}^{10}{\left( \dfrac{2}{sin\left( \dfrac{7\pi }{6}+k\pi \right)} \right)} \right]$
Or, ${{\sec }^{-1}}\left[ -\dfrac{1}{2}\sum\limits_{k=0}^{10}{\left( \dfrac{1}{sin\left( \dfrac{7\pi }{6}+k\pi \right)} \right)} \right]$
on solving, we get
${{\sec }^{-1}}\left[ -\dfrac{1}{2}\sum\limits_{k=0}^{10}{\left( \dfrac{1}{sin\left( \dfrac{\pi }{6}+k\pi +\pi \right)} \right)} \right]$
On re-arranging, we get
${{\sec }^{-1}}\left[ -\dfrac{1}{2}\sum\limits_{k=0}^{10}{\left( \dfrac{1}{sin\left( \dfrac{\pi }{6}+\pi (k+1) \right)} \right)} \right]$
We know that $\sin (k\pi +\theta )={{(-1)}^{k}}\sin \theta$
So, we can write $sin\left( \dfrac{\pi }{6}+\pi (k+1) \right)$, as ${{(-1)}^{(k+1)}}sin\left( \dfrac{\pi }{6} \right)$
Also, we know that $\sin \dfrac{\pi }{6}=\dfrac{1}{2}$
So, we can write ${{(-1)}^{(k+1)}}sin\left( \dfrac{\pi }{6} \right)$ as $\dfrac{{{(-1)}^{(k+1)}}}{2}$
So, we have ${{\sec }^{-1}}\left[ -\dfrac{1}{2}\sum\limits_{k=0}^{10}{\left( \dfrac{1}{\dfrac{{{(-1)}^{(k+1)}}}{2}} \right)} \right]$
On solving, we get
${{\sec }^{-1}}\left[ -\dfrac{1}{2}\sum\limits_{k=0}^{10}{\left( \dfrac{2}{{{(-1)}^{(k+1)}}} \right)} \right]$
Or, ${{\sec }^{-1}}\left[ -1\sum\limits_{k=0}^{10}{\left( \dfrac{1}{{{(-1)}^{(k+1)}}} \right)} \right]$
Now, on expanding $\sum\limits_{k=0}^{10}{\left( \dfrac{1}{{{(-1)}^{(k+1)}}} \right)}$, we get $\sum\limits_{k=0}^{10}{\left( \dfrac{1}{{{(-1)}^{(k+1)}}} \right)}=-1+1-1+1-1+1-1+1-1+1-1$
On simplifying, we get $\sum\limits_{k=0}^{10}{\left( \dfrac{1}{{{(-1)}^{(k+1)}}} \right)}=-1$
So, we have ${{\sec }^{-1}}\left[ -1\times -1 \right]$
Or, ${{\sec }^{-1}}\left[ 1 \right]$
We know that ${{\sec }^{-1}}\left[ 1 \right]=0$
So, value of ${{\sec }^{-1}}\left[ \dfrac{1}{4}\sum\limits_{k=0}^{10}{\left( \sec \left( \dfrac{7\pi }{12}+\dfrac{k\pi }{2} \right)\sec \left( \dfrac{7\pi }{12}+(k+1)\dfrac{\pi }{2} \right) \right)} \right]$ in the interval $\left[ -\dfrac{\pi }{4},\dfrac{3\pi }{4} \right]$ equals to 0.

Note: To, solve such question one must know the inverse trigonometric function properties such as ${{\sec }^{-1}}\left[ 1 \right]=0$ and trigonometric identities such as $\sec \left( \dfrac{\pi }{2}+\theta \right)=-\cos ec\theta$, $\sec x=\dfrac{1}{\cos x}$,$cosecx=\dfrac{1}{sinx}$ and very important $\sin (k\pi +\theta )={{(-1)}^{k}}\sin \theta$ and one must also know how to open summation function. Try not to make any silly mistake as this will make the final answer wrong.