JEE Advanced · Mathematics · 7. Trigonometry
The value of
\(\sec ^{-1}\left(\frac{1}{4} \sum_{k=0}^{10} \sec (\frac{7 \pi}{12}+\frac{k \pi}{2}\right) \sec (\frac{7 \pi}{12}\) \(+\frac{(k+1) \pi}{2}))\) in the interval \(\left[-\frac{\pi}{4}, \frac{3 \pi}{4}\right]\) equals
- A 0
- B 1
- C 2
- D 3
Answer & Solution
Correct Answer
(A) 0
Step-by-step Solution
Detailed explanation
\(\sec ^{-1}[\frac{1}{4} \sum_{k=0}^{10} \sec \left(\frac{7 \pi}{12}+\frac{k \pi}{2}\right) \sec (\frac{7 \pi}{12}~+\) \(\frac{k \pi}{2}+\frac{\pi}{2})] \)
\( =\sec ^{-1}[\frac{-1}{4} \sum_{k=0}^{10} \sec \left(\frac{7 \pi}{12}+\frac{k \pi}{2}\right)\) \(\operatorname{cosec}\left(\frac{7 \pi}{12}+\frac{k \pi}{2}\right)] \quad\) \(\left(\therefore \sec \left(\frac{\pi}{2}+\theta\right)=-\operatorname{cosec} \theta\right) \)
\( =\sec ^{-1}\left(-\frac{1}{4} \sum_{k=0}^{10} \frac{1}{\cos \left(\frac{7 \pi}{12}+\frac{k \pi}{2}\right) \sin \left(\frac{7 \pi}{12}+\frac{k \pi}{2}\right)}\right) \)
\( =\sec ^{-1}\left(-\frac{1}{4} \sum_{k=0}^{10} \frac{2}{\sin \left(\frac{7 \pi}{6}+k \pi\right)}\right)(\therefore\) \(2 \sin \theta \cos \theta=\sin 2 \theta)\)
now if \(k=\) even \(\sin \left(2 n \pi+\frac{7 \pi}{6}\right)=\sin \frac{7 \pi}{6}=-\frac{1}{2}\)
and if \(k=\) odd \(\sin \left((2 n+1) \pi+\frac{7 \pi}{6}\right)=-\sin \frac{7 \pi}{6}=\frac{1}{2}\)
hence
\(=\sec ^{-1}\left(-\frac{1}{2}\left(\frac{1}{-\frac{1}{2}}+\frac{1}{\frac{1}{2}}+\frac{1}{-\frac{1}{2}}+\ldots \ldots\right)\right) \)
\( =\sec ^{-1} 1 \)
\( =0\)
\( =\sec ^{-1}[\frac{-1}{4} \sum_{k=0}^{10} \sec \left(\frac{7 \pi}{12}+\frac{k \pi}{2}\right)\) \(\operatorname{cosec}\left(\frac{7 \pi}{12}+\frac{k \pi}{2}\right)] \quad\) \(\left(\therefore \sec \left(\frac{\pi}{2}+\theta\right)=-\operatorname{cosec} \theta\right) \)
\( =\sec ^{-1}\left(-\frac{1}{4} \sum_{k=0}^{10} \frac{1}{\cos \left(\frac{7 \pi}{12}+\frac{k \pi}{2}\right) \sin \left(\frac{7 \pi}{12}+\frac{k \pi}{2}\right)}\right) \)
\( =\sec ^{-1}\left(-\frac{1}{4} \sum_{k=0}^{10} \frac{2}{\sin \left(\frac{7 \pi}{6}+k \pi\right)}\right)(\therefore\) \(2 \sin \theta \cos \theta=\sin 2 \theta)\)
now if \(k=\) even \(\sin \left(2 n \pi+\frac{7 \pi}{6}\right)=\sin \frac{7 \pi}{6}=-\frac{1}{2}\)
and if \(k=\) odd \(\sin \left((2 n+1) \pi+\frac{7 \pi}{6}\right)=-\sin \frac{7 \pi}{6}=\frac{1}{2}\)
hence
\(=\sec ^{-1}\left(-\frac{1}{2}\left(\frac{1}{-\frac{1}{2}}+\frac{1}{\frac{1}{2}}+\frac{1}{-\frac{1}{2}}+\ldots \ldots\right)\right) \)
\( =\sec ^{-1} 1 \)
\( =0\)
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