Verified Solution • 2025–26 Session moderate

QJEEM17Mad4e0: JEE Main 2017 Mathematics - Class 12 JEEM17Mad4e0 Solution: The value of tan -1 is equal to :

The value of tan-1 [1+x2+1−x21+x2−1−x2],\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right], \left| x \right| < {1 \over 2},x \ne 0, is equal to :

Multiple Choice Options

  • A π4+12cos⁡−1 x2{\pi \over 4} + {1 \over 2}{\cos ^{ - 1}}\,{x^2}
  • B π4+cos⁡−1 x2{\pi \over 4} + {\cos ^{ - 1}}\,{x^2}
  • C π4−12cos⁡−1 x2{\pi \over 4} - {1 \over 2}{\cos ^{ - 1}}\,{x^2}
  • D π4−cos⁡−1 x2{\pi \over 4} - {\cos ^{ - 1}}\,{x^2}

Step-by-Step Verified Solution

Given,

tan-1 [1+x2+1−x21+x2−1−x2]\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right]

Let    x2 = cos θ\theta

= tan-1 [1+cos⁡θ+1−cos⁡θ1+cos⁡θ−1−cos⁡θ]\left[ {{{\sqrt {1 + \cos \theta } + \sqrt {1 - \cos \theta } } \over {\sqrt {1 + \cos \theta } - \sqrt {1 - \cos \theta } }}} \right]

= tan-1 [2cos⁡2θ2+2sin⁡2θ22cos⁡2θ2−2sin⁡2θ2]\left[ {{{\sqrt {2{{\cos }^2}{\theta \over 2}} + \sqrt {2{{\sin }^2}{\theta \over 2}} } \over {\sqrt {2{{\cos }^2}{\theta \over 2}} - \sqrt {2{{\sin }^2}{\theta \over 2}} }}} \right]

= tan-1 [2cos⁡θ2+2sin⁡θ22cos⁡θ2−2sin⁡θ2]\left[ {{{\sqrt 2 \cos {\theta \over 2} + \sqrt 2 \sin {\theta \over 2}} \over {\sqrt 2 \cos {\theta \over 2} - \sqrt 2 \sin {\theta \over 2}}}} \right]

= tan-1 [cos⁡θ2+sin⁡θ2cos⁡θ2−sin⁡θ2]\left[ {{{\cos {\theta \over 2} + \sin {\theta \over 2}} \over {\cos {\theta \over 2} - \sin {\theta \over 2}}}} \right]

= tan-1 [1+tan⁡θ21−tan⁡θ2]\left[ {{{1 + \tan {\theta \over 2}} \over {1 - \tan {\theta \over 2}}}} \right]

= tan-1 [tan⁡π4+tan⁡θ21−tan⁡π4+tan⁡θ2]\left[ {{{\tan {\pi \over 4} + \tan {\theta \over 2}} \over {1 - \tan {\pi \over 4} + \tan {\theta \over 2}}}} \right]

= tan-1 (tan⁡(π4+θ2))\left( {\tan \left( {{\pi \over 4} + {\theta \over 2}} \right)} \right)

= π4+θ2{\pi \over 4} + {\theta \over 2}

= π4+12{\pi \over 4} + {1 \over 2} cos-1 x2

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