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Answer 1

Here,

$\begin{aligned}\displaystyle\longrightarrow\ \ &\sf{\sec^{-1}\left(\dfrac{1}{4}\sum_{k=0}^{10}\sec\left(\dfrac{7\pi}{12}+\dfrac{k\pi}{2}\right)\sec\left(\dfrac{7\pi}{12}+\dfrac{(k+1)\pi}{2}\right)\right)}\\\\=\ \ &\sf{\sec^{-1}\left(\dfrac{1}{4}\sum_{k=0}^{10}\sec\left(\dfrac{7\pi}{12}+\dfrac{k\pi}{2}\right)\sec\left(\dfrac{7\pi}{12}+\dfrac{k\pi}{2}+\dfrac{\pi}{2}\right)\right)}\end{aligned}$

Since $\sf{\sec\left(\dfrac{\pi}{2}+x\right)=-\csc x,}$

$\begin{aligned}\displaystyle\longrightarrow\ \ &\sf{\sec^{-1}\left(\dfrac{1}{4}\sum_{k=0}^{10}\sec\left(\dfrac{7\pi}{12}+\dfrac{k\pi}{2}\right)\sec\left(\dfrac{7\pi}{12}+\dfrac{(k+1)\pi}{2}\right)\right)}\\\\=\ \ &\sf{\sec^{-1}\left(-\dfrac{1}{4}\sum_{k=0}^{10}\sec\left(\dfrac{7\pi}{12}+\dfrac{k\pi}{2}\right)\csc\left(\dfrac{7\pi}{12}+\dfrac{k\pi}{2}\right)\right)}\end{aligned}$

Taking $\sf{\sec x=\dfrac{1}{\cos x}}$ and $\sf{\csc x=\dfrac{1}{\sin x},}$

$\begin{aligned}\displaystyle\longrightarrow\ \ &\sf{\sec^{-1}\left(\dfrac{1}{4}\sum_{k=0}^{10}\sec\left(\dfrac{7\pi}{12}+\dfrac{k\pi}{2}\right)\sec\left(\dfrac{7\pi}{12}+\dfrac{(k+1)\pi}{2}\right)\right)}\\\\=\ \ &\sf{\sec^{-1}\left(-\dfrac{1}{4}\sum_{k=0}^{10}\dfrac{1}{\sin\left(\dfrac{7\pi}{12}+\dfrac{k\pi}{2}\right)\cos\left(\dfrac{7\pi}{12}+\dfrac{k\pi}{2}\right)}\right)}\end{aligned}$

$\begin{aligned}\displaystyle\longrightarrow\ \ &\sf{\sec^{-1}\left(\dfrac{1}{4}\sum_{k=0}^{10}\sec\left(\dfrac{7\pi}{12}+\dfrac{k\pi}{2}\right)\sec\left(\dfrac{7\pi}{12}+\dfrac{(k+1)\pi}{2}\right)\right)}\\\\=\ \ &\sf{\sec^{-1}\left(-\dfrac{1}{2}\sum_{k=0}^{10}\dfrac{1}{2\sin\left(\dfrac{7\pi}{12}+\dfrac{k\pi}{2}\right)\cos\left(\dfrac{7\pi}{12}+\dfrac{k\pi}{2}\right)}\right)}\end{aligned}$

Since $\sf{2\sin x\cos x=\sin(2x),}$

$\begin{aligned}\displaystyle\longrightarrow\ \ &\sf{\sec^{-1}\left(\dfrac{1}{4}\sum_{k=0}^{10}\sec\left(\dfrac{7\pi}{12}+\dfrac{k\pi}{2}\right)\sec\left(\dfrac{7\pi}{12}+\dfrac{(k+1)\pi}{2}\right)\right)}\\\\=\ \ &\sf{\sec^{-1}\left(-\dfrac{1}{2}\sum_{k=0}^{10}\dfrac{1}{\sin\left(\dfrac{7\pi}{6}+k\pi\right)}\right)}\end{aligned}$

$\begin{aligned}\displaystyle\longrightarrow\ \ &\sf{\sec^{-1}\left(\dfrac{1}{4}\sum_{k=0}^{10}\sec\left(\dfrac{7\pi}{12}+\dfrac{k\pi}{2}\right)\sec\left(\dfrac{7\pi}{12}+\dfrac{(k+1)\pi}{2}\right)\right)}&\\\\=\ \ &\sf{\sec^{-1}\left(-\dfrac{1}{2}\sum_{k=0}^{10}\dfrac{1}{\sin\left((k+1)\pi+\dfrac{\pi}{6}\right)}\right)}&\sf{\quad\quad\dots(1)}\end{aligned}$

If $\sf{k}$ is even, $\sf{\sin\left((k+1)\pi+\dfrac{\pi}{6}\right)=-\sin\left(\dfrac{\pi}{6}\right)}$

If $\sf{k}$ is odd, $\sf{\sin\left((k+1)\pi+\dfrac{\pi}{6}\right)=\sin\left(\dfrac{\pi}{6}\right)}$

Hence we see that,

$\displaystyle\longrightarrow\sf{\sum_{k=0}^{10}\dfrac{1}{\sin\left((k+1)\pi+\dfrac{\pi}{6}\right)}=\sum_{k=0}^{10}\dfrac{(-1)^{k+1}}{\sin\left(\dfrac{\pi}{6}\right)}}$

$\displaystyle\longrightarrow\sf{\sum_{k=0}^{10}\dfrac{1}{\sin\left((k+1)\pi+\dfrac{\pi}{6}\right)}=\underbrace{\sf{-\dfrac{1}{\sin\left(\dfrac{\pi}{6}\right)}+\dfrac{1}{\sin\left(\dfrac{\pi}{6}\right)}-\,\dots\,-\dfrac{1}{\sin\left(\dfrac{\pi}{6}\right)}}}_{11\ terms}}$

$\displaystyle\longrightarrow\sf{\sum_{k=0}^{10}\dfrac{1}{\sin\left((k+1)\pi+\dfrac{\pi}{6}\right)}=-\dfrac{1}{\sin\left(\dfrac{\pi}{6}\right)}}$

$\displaystyle\longrightarrow\sf{\sum_{k=0}^{10}\dfrac{1}{\sin\left((k+1)\pi+\dfrac{\pi}{6}\right)}=-2}$

Thus (1) becomes,

$\displaystyle\longrightarrow\sf{\sec^{-1}\left(\dfrac{1}{4}\sum_{k=0}^{10}\sec\left(\dfrac{7\pi}{12}+\dfrac{k\pi}{2}\right)\sec\left(\dfrac{7\pi}{12}+\dfrac{(k+1)\pi}{2}\right)\right)=\sec^{-1}\left(-\dfrac{1}{2}\times-2\right)}$

$\displaystyle\longrightarrow\sf{\sec^{-1}\left(\dfrac{1}{4}\sum_{k=0}^{10}\sec\left(\dfrac{7\pi}{12}+\dfrac{k\pi}{2}\right)\sec\left(\dfrac{7\pi}{12}+\dfrac{(k+1)\pi}{2}\right)\right)=\sec^{-1}(1)}$

In the interval $\sf{\left[-\dfrac{\pi}{4},\ \dfrac{3\pi}{4}\right],}$

$\displaystyle\longrightarrow\underline{\underline{\sf{\sec^{-1}\left(\dfrac{1}{4}\sum_{k=0}^{10}\sec\left(\dfrac{7\pi}{12}+\dfrac{k\pi}{2}\right)\sec\left(\dfrac{7\pi}{12}+\dfrac{(k+1)\pi}{2}\right)\right)=0}}}$

Hence 0 is the answer.

Answer 2

$\bf The\:\:value\:\: of$

$\bf {sec}^{-1} \bigg[ \dfrac{1}{4} \sum \limits_{k=0}^{10}\sec \big( \dfrac{7 \pi}{12}+ \dfrac{k \pi}{2} \big) sec \big( \dfrac{7 \pi}{12}+ \dfrac{(k+1) \pi}{2} \big) \bigg] \\ \\ \\ \\ \bf in\:\:the\:\:interval \bigg[ -\dfrac{\pi}{4} ,\dfrac{3 \pi}{4} \bigg] equals---$

$\rule{200}2$

$\red{\underline{\underline{\sf{SOLUTION}}}}$

$\rule{200}2$

$\bf \sum \limits_{k=0}^{10} \:\: \dfrac{ \huge{1} }{ \bigg[ cos(\frac{7 \pi}{12}+\frac{k \pi}{2})(cos \frac{7 \pi}{12}+\frac{k \pi}{2}+ \frac{\pi}{2} ) \bigg] }$

$\implies \bf \sum \limits_{k=0}^{10} \:\: \dfrac{ sin \bigg[(\frac{7 \pi}{12}+\frac{k \pi}{2}+ \frac{\pi}{2})-(\frac{7 \pi}{12}+\frac{k \pi}{2}) \bigg] }{ \bigg[ cos(\frac{7 \pi}{12}+\frac{k \pi}{2})\:cos (\frac{7 \pi}{12}+\frac{k \pi}{2}+ \frac{\pi}{2} ) \bigg] }$

We know that,

$\bf sin(A-B)=sinA cosB - cos A sinB$

$\implies \bf \sum \limits_{k=0}^{10} \:\: \dfrac{ \bigg[sin(\frac{7 \pi}{12}+\frac{k \pi}{2}+ \frac{\pi}{2}) cos(\frac{7 \pi}{12}+\frac{k \pi}{2})- cos(\frac{7 \pi}{12}+\frac{k \pi}{2}+ \frac{\pi}{2}) sin(\frac{7 \pi}{12}+\frac{k \pi}{2}) \bigg] }{ \bigg[ cos(\frac{7 \pi}{12}+\frac{k \pi}{2})\:cos (\frac{7 \pi}{12}+\frac{k \pi}{2}+ \frac{\pi}{2} ) \bigg] }$

$\implies \bf \sum \limits_{k=0}^{10} \:\:\:\:\:\dfrac{sin(\frac{7 \pi}{12}+\frac{k \pi}{2}+ \frac{\pi}{2}) \cancel{cos(\frac{7 \pi}{12}+\frac{k \pi}{2}) } }{ \cancel{cos(\frac{7 \pi}{12}+\frac{k \pi}{2}) }\times cos (\frac{7 \pi}{12}+\frac{k \pi}{2}+ \frac{\pi}{2} )} - \dfrac{ \cancel{cos(\frac{7 \pi}{12}+\frac{k \pi}{2}+ \frac{\pi}{2})} sin(\frac{7 \pi}{12}+\frac{k \pi}{2})}{ cos(\frac{7 \pi}{12}+\frac{k \pi}{2})\times \cancel{cos (\frac{7 \pi}{12}+\frac{k \pi}{2}+ \frac{\pi}{2} )}}$

$\implies \bf \sum \limits_{k=0}^{10} \:\: tan \big( \frac{7 \pi}{12}+\frac{k \pi}{2} +\frac{\pi}{2} \big) - tan \big( \frac{7 \pi}{12}+\frac{k \pi}{2} \big)$

$\implies \bf tan \big( \frac{7 \pi}{12}+\frac{11 \pi}{2} \big) - tan \big( \frac{7 \pi}{12}\big) \:\:\: [\bf \underline{Summation\:\:of\:\:series}]$

$\implies \bf tan \bigg( \dfrac{6 \pi}{12}+\dfrac{\pi}{12}+\dfrac{11 \pi}{2} \bigg) - tan \bigg( \dfrac{6 \pi}{12}+\dfrac{\pi}{12}\bigg)$

$\implies \bf tan \bigg( \dfrac{\pi}{12}+\dfrac{11\pi}{2}+\dfrac{ \pi}{12} \bigg) - tan \bigg( \dfrac{ \pi}{2}+\dfrac{\pi}{12}\bigg)$

$\implies \bf tan \bigg( \dfrac{12 \pi}{2}+\dfrac{\pi}{12} \bigg) - tan \bigg( \dfrac{ \pi}{2}+\dfrac{\pi}{12}\bigg)$

$\implies \bf tan (6\pi+\frac{\pi}{12})-tan(\frac{\pi}{2}+\frac{\pi}{12})$

$\implies \bf tan (\frac{\pi}{12})-[-cot(\frac{\pi}{12})\:]$

$\implies \bf tan (\frac{\pi}{12})+cot(\frac{\pi}{12})$

$\implies \bf \dfrac{sin \frac{\pi}{12}}{cos \frac{\pi}{12}}+\dfrac{cos \frac{\pi}{12}}{sin \frac{\pi}{12}}$

$\implies \bf \dfrac{sin^{2} \frac{\pi}{12}+cos^{2} \frac{\pi}{12}}{cos \frac{\pi}{12}\times sin \frac{\pi}{12}}$

$\implies \bf \dfrac{1}{cos \frac{\pi}{12}\times sin \frac{\pi}{12}}$

$\implies \bf \dfrac{2}{2\:sin \frac{\pi}{12} cos \frac{\pi}{12}} = \dfrac{2}{sin(2 \times \frac{\pi}{12})}= \dfrac{2}{sin \frac{\pi}{6}} = \frac{2}{\frac{1}{2}}=4$

$\implies \bf sec^{-1} \:\bigg[\frac{1}{4} \sum \limits_{k=0}^{10}\sec(\frac{7\pi}{12}+\frac{k\pi}{2})sec(\frac{7\pi}{12}+(\frac{k+1\pi}{2}) \big) \bigg]$

$\implies \bf sec^{-1} (\frac{1}{4} \times 4)$

$\implies \bf sec^{-1}(1)$

$\boxed{\boxed{\implies \bf sec^{-1}(1)=0}}$

$\rule{200}2$

Hence, answer is 0.