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CBSE Class 12 Math 2022 Term I Solved Paper

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Question : 19 of 50
Marks: +1, -0
The principal value of tan⁡−1(tan⁡9π8)\tan^{-1}\left(\tan \frac{9\pi}{8}\right) is
Solution:     👈: Video Solution
Explanation: tan⁡−1(tan⁡9π8)=tan⁡−1(tan⁡(π+π8))\tan^{-1}\left(\tan \frac{9\pi}{8}\right) = \tan^{-1}\left(\tan \left(\pi + \frac{\pi}{8}\right)\right)
=tan⁡−1(tan⁡π8)=π8[∵π8∈(−π2,π2)]= \tan^{-1}\left(\tan \frac{\pi}{8}\right) = \frac{\pi}{8} \quad \left[ \because \frac{\pi}{8} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \right]
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