Verified Solution • 2025–26 Session moderate

QJEEM19Madd0e: JEE Main 2019 Mathematics - Class 12 JEEM19Madd0e Solution: If , where , then - is equal to :

If α=cos⁡−1(35)\alpha = {\cos ^{ - 1}}\left( {{3 \over 5}} \right) , β=tan⁡−1(13)\beta = {\tan ^{ - 1}}\left( {{1 \over 3}} \right) where 0 < \alpha ,\beta < {\pi \over 2} , then α\alpha - β\beta is equal to :

Multiple Choice Options

  • A tan⁡−1(914){\tan ^{ - 1}}\left( {{9 \over {14 }}} \right)
  • B sin⁡−1(9510){\sin ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)
  • C cos⁡−1(9510){\cos ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)
  • D tan⁡−1(9510){\tan ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)

Step-by-Step Verified Solution

Here cos⁡α=35\cos \alpha = {3 \over 5}

∴\therefore tan⁡α=43\tan \alpha = {4 \over 3}

and tan⁡β=13\tan \beta = {1 \over 3}

We know,

tan⁡(α−β)=tan⁡α−tan⁡β1+tan⁡α.tan⁡β\tan \left( {\alpha - \beta } \right) = {{\tan \alpha - \tan \beta } \over {1 + \tan \alpha .\tan \beta }}

= 43−131+43.13{{{4 \over 3} - {1 \over 3}} \over {1 + {4 \over 3}.{1 \over 3}}} = 913{9 \over {13}}

∴\therefore (α−β)\left( {\alpha - \beta } \right) = tan⁡−1(913){\tan ^{ - 1}}\left( {{9 \over {13}}} \right) = sin⁡−1(9510){\sin ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)

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