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  • need solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 6 sub question (v)

need solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 6 sub question (v)

Answers (1)

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Answer: \frac{\pi }{3}

Hint: The range of principal value of    \cot ^{-1} is \left [ 0,\pi \right ]
Given:  \cot ^{-1}\left\{\cot \left(-\frac{8 \pi}{3}\right)\right\}

Explanation:

First we solve  \cot \left(-\frac{8 \pi}{3}\right)

As we know  \cot (-\theta)=-\cot \theta

            \begin{aligned} &\cot \left(-\frac{8 \pi}{3}\right)=-\cot \left(\frac{8 \pi}{3}\right) \\ &-\cot \left(\frac{8 \pi}{3}\right)=-\cot \left(3 \pi-\frac{\pi}{3}\right) \end{aligned}

As we know, \cot (2 \pi-\theta)=-\cot \theta

            \begin{aligned} -\cot \left(3 \pi-\frac{\pi}{3}\right) &=-\cot \left(-\frac{\pi}{3}\right) \\ &=\cot \left(\frac{\pi}{3}\right) \\ \cot \left(\frac{\pi}{3}\right) &=\frac{1}{\sqrt{3}} \end{aligned}

By substituting these values in \cot ^{-1}\left(\cot \left(-\frac{8 \pi}{3}\right)\right) we get,

            \cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)

Now,

Let       y=\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)

          \begin{aligned} &\cot y=\frac{1}{\sqrt{3}} \\ &\cot \left(\frac{\pi}{3}\right)=\frac{1}{\sqrt{3}} \end{aligned}

The range of the principal value of  \cot ^{-1} \text { is }[0, \pi]

         \begin{aligned} &\cot ^{-1}\left(\cot \frac{\pi}{3}\right)=\frac{\pi}{3} \\ &\therefore \cot ^{-1}(\cot x)=x, \text { where } x \in[0, \pi] \end{aligned}

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