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  • explain solution RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 6 sub question (vi) maths

explain solution RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 6 sub question (vi) maths

Answers (1)

Answer: \frac{\pi }{4}

Hint: The range of principal value of    \cot ^{-1} is \left [ 0,\pi \right ]
Given:  \cot ^{-1}\left\{\cot \left(\frac{21 \pi}{4}\right)\right\}

Explanation:

First we solve \cot \left(\frac{21 \pi}{4}\right)

            \cot \left(\frac{21 \pi}{4}\right)=\cot \left(5 \pi+\frac{\pi}{4}\right)

As we know  \cot (n \pi+\theta)=\cot \theta \text { where } n \text { is }(1,2,3, \ldots)

            \cot \left(5 \pi+\frac{\pi}{4}\right)=\cot \left(\frac{\pi}{4}\right) \quad\left[\because \cot \frac{\pi}{4}=1\right]

            \cot \left(\frac{\pi}{4}\right)=1

By substituting these values in  \cot ^{-1}\left\{\cot \left(\frac{21 \pi}{4}\right)\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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