Verified Solution • 2025–26 Session moderate

QJEEM22Ma1656: JEE Main 2022 Mathematics - Class 12 JEEM22Ma1656 Solution

sin⁡1(sin⁡2π3)+cos⁡−1(cos⁡7π6)+tan⁡−1(tan⁡3π4){\sin ^1}\left( {\sin {{2\pi } \over 3}} \right) + {\cos ^{ - 1}}\left( {\cos {{7\pi } \over 6}} \right) + {\tan ^{ - 1}}\left( {\tan {{3\pi } \over 4}} \right) is equal to :

Multiple Choice Options

  • A 11π12{{11\pi } \over {12}}
  • B 17π12{{17\pi } \over {12}}
  • C 31π12{{31\pi } \over {12}}
  • D −- 3π4{{3\pi } \over {4}}

Step-by-Step Verified Solution

sin⁡−1(32)+cos⁡−1(−32)+tan⁡−1(−1){\sin ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} \right) + {\cos ^{ - 1}}\left( {{{ - \sqrt 3 } \over 2}} \right) + {\tan ^{ - 1}}\left( { - 1} \right)

=π3+5π6−π4= {\pi \over 3} + {{5\pi } \over 6} - {\pi \over 4}

=4π+10π−3π12=11π12= {{4\pi + 10\pi - 3\pi } \over {12}} = {{11\pi } \over {12}}

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