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Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 6 sub question (iii) maths textbook solution

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Answer: \frac{\pi }{4}

Hint: The range of principal value of    \cot ^{-1} is \left [ 0,\pi \right ]
Given:  \cot ^{-1}\left(\cot \frac{9\pi}{4}\right)


First we solve  \cot \frac{9 \pi}{4}

            \begin{aligned} &\cot \frac{9 \pi}{4}=\cot \left(2 \pi+\frac{\pi}{4}\right) \\ &\therefore \cot (2 \pi+\theta)=\cot \theta \\ &\cot \left(2 \pi+\frac{\pi}{4}\right)=\cot \frac{\pi}{4} \\ &\therefore \cot \frac{\pi}{4}=1 \end{aligned}

By substituting these value in \cot ^{-1}\left(\cot \frac{9 \pi}{4}\right) we get,

            \cot ^{-1}(1) 

Now,  \text { let } y=\cot ^{-1}(1)

            \begin{aligned} &\cot y=1 \\ &\cot \left(\frac{\pi}{4}\right)=1 \end{aligned}

The range of principal value of     \cot ^{-1} \text { is }[0, \pi] \text { and } \cot \left(\frac{\pi}{4}\right)=1

            \begin{aligned} &\therefore \cot ^{-1}\left(\cot \frac{\pi}{4}\right)=\frac{\pi}{4} \\ &\cot ^{-1}(\cot x)=x, x \in[0, \pi] \end{aligned}

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