Verified Solution • 2025–26 Session moderate

QJEEM22Maffad: JEE Main 2022 Mathematics - Class 12 JEEM22Maffad Solution

The value of tan⁡−1(cos⁡(15π4)−1sin⁡(π4)){\tan ^{ - 1}}\left( {{{\cos \left( {{{15\pi } \over 4}} \right) - 1} \over {\sin \left( {{\pi \over 4}} \right)}}} \right) is equal to :

Multiple Choice Options

  • A −π4- {\pi \over 4}
  • B −π8- {\pi \over 8}
  • C −5π12- {{5\pi } \over {12}}
  • D −4π9- {{4\pi } \over 9}

Step-by-Step Verified Solution

tan⁡−1(cos⁡(15π4)−1sin⁡π4){\tan ^{ - 1}}\left( {{{\cos \left( {{{15\pi } \over 4}} \right) - 1} \over {\sin {\pi \over 4}}}} \right)

=tan⁡−1(12−112)= {\tan ^{ - 1}}\left( {{{{1 \over {\sqrt 2 }} - 1} \over {{1 \over {\sqrt 2 }}}}} \right)

=tan⁡−1(1−2)=−tan⁡−1(2−1)= {\tan ^{ - 1}}(1 - \sqrt 2 ) = - {\tan ^{ - 1}}(\sqrt 2 - 1)

=−π8= - {\pi \over 8}

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