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

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

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Answer: \frac{\pi }{6}

Hint: The range of principal value of    \cot ^{-1} is \left [ 0,\pi \right ]
Given:  \cot ^{-1}\left(\cot \frac{19\pi}{6}\right)

Explanation:

First we solve  \cot \left(\frac{19 \pi}{6}\right)

            \cot \left(\frac{19 \pi}{6}\right)=\cot \left(3 \pi+\frac{\pi}{6}\right)

As we know,  \cot (n \pi+\theta)=\cot \theta

            \begin{aligned} &\cot \left(3 \pi+\frac{\pi}{6}\right)=\cot \frac{\pi}{6} \\ &\cot \frac{\pi}{6}=\sqrt{3} \end{aligned}                                \left[\because \cot 30^{\circ}=\sqrt{3}\right]

By substituting these value in \cot ^{-1}\left(\cot \frac{19 \pi}{6}\right)    we get,

            \cot ^{-1}(\sqrt{3})

Now,     \text { let } y=\cot ^{-1}(\sqrt{3})

            \begin{aligned} &\cot y=\sqrt{3} \\ &\cot \left(\frac{\pi}{6}\right)=\sqrt{3} \end{aligned}

The range of the principal value of   \cot ^{-1} \text { is }[0, \pi] \text { and } \cot \left(\frac{\pi}{6}\right)=\sqrt{3}

            \therefore \cot ^{-1}\left(\cot \frac{\pi}{6}\right)=\frac{\pi}{6}

As we know, \cot ^{-1}(\cot x)=x, x \in[0, \pi]

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