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  • need solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 4 sub question (v)

need solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 4 sub question (v)

Answers (1)

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Answer:  \frac{\pi }{5}

Hint: The range of principal value of  \sec ^{-1} is \left [ 0,\pi \right ]-\left [ \frac{\pi }{2} \right ]
Given:  \sec ^{-1}\left(\sec \frac{9 \pi}{5}\right)

Explanation:

First we solve \sec \left(\frac{9 \pi}{5}\right)

            \begin{aligned} &\sec \left(\frac{9 \pi}{5}\right)=\sec \left(2 \pi-\frac{\pi}{5}\right) \\ &\therefore[\sec (2 \pi-\theta)]=\sec \theta \end{aligned}

            \begin{aligned} &\sec \left(2 \pi-\frac{\pi}{5}\right)=\sec \left(\frac{\pi}{5}\right) \\ &\sec \left(\frac{9 \pi}{5}\right)=\sec \left(\frac{\pi}{5}\right) \end{aligned}

By substituting these value in \sec ^{-1}\left(\sec \frac{9 \pi}{5}\right) we get,

            \begin{aligned} &\sec ^{-1}\left(\sec \frac{\pi}{5}\right) \\ &\therefore \sec ^{-1}(\sec x)=x, x \in[0, \pi]-\left\{\frac{\pi}{2}\right\} \\ &\sec ^{-1}\left(\sec \frac{\pi}{5}\right)=\frac{\pi}{5} \end{aligned}

Hence, \sec ^{-1}\left(\sec \frac{9 \pi}{5}\right)=\frac{\pi}{5}

 

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