Find the principal value of \(\tan^{-1}\left(\tan\left(-\dfrac{5\pi}{6}\right)\right)\).\(\tan^{-1}\left(\tan\left(-\dfrac{5\pi}{6}\right)\right)\) का principal value (प्रधान मान) ज्ञात कीजिए।
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(\pi/6\)\(\pi/6\)
The principal value of \(\tan^{-1}(y)\) lies in \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\). Now \(\tan\left(-\dfrac{5\pi}{6}\right)=\tan\left(-\dfrac{5\pi}{6}+\pi\right)=\tan\left(\dfrac{\pi}{6}\right)\). Therefore the angle whose tangent equals \(\tan\left(-\dfrac{5\pi}{6}\right)\) and lies in \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\) is \(\dfrac{\pi}{6}\). Hence the required value is \(\dfrac{\pi}{6}\).\(\tan^{-1}(y)\) का principal value \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\) के बीच होता है। अब \(\tan\left(-\dfrac{5\pi}{6}\right)=\tan\left(-\dfrac{5\pi}{6}+\pi\right)=\tan\left(\dfrac{\pi}{6}\right)\). अतः जिसका \(\tan\) \(\tan\left(-\dfrac{5\pi}{6}\right)\) के बराबर है और जो \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\) में आता है, वह कोण \(\dfrac{\pi}{6}\) है। इसलिए वांछित मान \(\dfrac{\pi}{6}\) है।
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