Verified Solution • 2025–26 Session moderate

QJEEM20Ma332c: JEE Main 2020 Mathematics - Class 12 JEEM20Ma332c Solution: 2 - is equal to :

2 π\pi - (sin⁡−145+sin⁡−1513+sin⁡−11665)\left( {{{\sin }^{ - 1}}{4 \over 5} + {{\sin }^{ - 1}}{5 \over {13}} + {{\sin }^{ - 1}}{{16} \over {65}}} \right) is equal to :

Multiple Choice Options

  • A 7π4{{7\pi } \over 4}
  • B 5π4{{5\pi } \over 4}
  • C 3π2{{3\pi } \over 2}
  • D π2{\pi \over 2}

Step-by-Step Verified Solution

2π−(sin⁡−145+sin⁡−1513+sin⁡−11665)2\pi - \left( {{{\sin }^{ - 1}}{4 \over 5} + {{\sin }^{ - 1}}{5 \over {13}} + {{\sin }^{ - 1}}{{16} \over {65}}} \right)

=2π−(tan⁡−143+tan⁡−1512+tan⁡−11663)= 2\pi - \left( {{{\tan }^{ - 1}}{4 \over 3} + {{\tan }^{ - 1}}{5 \over {12}} + {{\tan }^{ - 1}}{{16} \over {63}}} \right)

=2π−{tan⁡−1(43+5121−43.512)+tan⁡−11663}= 2\pi - \left\{ {{{\tan }^{ - 1}}\left( {{{{4 \over 3} + {5 \over {12}}} \over {1 - {4 \over 3}.{5 \over {12}}}}} \right) + {{\tan }^{ - 1}}{{16} \over {63}}} \right\}

=2π−(tan⁡−16316+tan⁡−11663)= 2\pi - \left( {{{\tan }^{ - 1}}{{63} \over {16}} + {{\tan }^{ - 1}}{{16} \over {63}}} \right)

= 2π−(tan⁡−16316+cot⁡−16316)2\pi - \left( {{{\tan }^{ - 1}}{{63} \over {16}} + {{\cot }^{ - 1}}{{63} \over {16}}} \right)

=2π−π2= 2\pi - {\pi \over 2}

=3π2= {{3\pi } \over 2}

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