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  • explain solution RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 4 sub question (ii) maths

explain solution RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 4 sub question (ii) maths

Answers (1)

Answer: \frac{2\pi }{3}

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{2 \pi}{3}\right)

Explanation:

As         \sec ^{-1}(\sec x)=x \text { if } x \in[0, \pi]-\left\{\frac{\pi}{2}\right\}

By applying this situation we get,

             \sec ^{-1}\left(\sec \frac{2 \pi}{3}\right)=\frac{2 \pi}{3}

Hence,  \sec ^{-1}\left(\sec \frac{2 \pi}{3}\right)=\frac{2 \pi}{3}

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