JEE MAIN - Mathematics (2017 - 8th April Morning Slot - No. 10)
The value of tan-1 $$\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right],$$ $$\left| x \right| < {1 \over 2},x \ne 0,$$ is equal to :
$${\pi \over 4} + {1 \over 2}{\cos ^{ - 1}}\,{x^2}$$
$${\pi \over 4} + {\cos ^{ - 1}}\,{x^2}$$
$${\pi \over 4} - {1 \over 2}{\cos ^{ - 1}}\,{x^2}$$
$${\pi \over 4} - {\cos ^{ - 1}}\,{x^2}$$
Explanation
Given,
tan-1 $$\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right]$$
Let x2 = cos $$\theta $$
= tan-1 $$\left[ {{{\sqrt {1 + \cos \theta } + \sqrt {1 - \cos \theta } } \over {\sqrt {1 + \cos \theta } - \sqrt {1 - \cos \theta } }}} \right]$$
= tan-1 $$\left[ {{{\sqrt {2{{\cos }^2}{\theta \over 2}} + \sqrt {2{{\sin }^2}{\theta \over 2}} } \over {\sqrt {2{{\cos }^2}{\theta \over 2}} - \sqrt {2{{\sin }^2}{\theta \over 2}} }}} \right]$$
= tan-1 $$\left[ {{{\sqrt 2 \cos {\theta \over 2} + \sqrt 2 \sin {\theta \over 2}} \over {\sqrt 2 \cos {\theta \over 2} - \sqrt 2 \sin {\theta \over 2}}}} \right]$$
= tan-1 $$\left[ {{{\cos {\theta \over 2} + \sin {\theta \over 2}} \over {\cos {\theta \over 2} - \sin {\theta \over 2}}}} \right]$$
= tan-1 $$\left[ {{{1 + \tan {\theta \over 2}} \over {1 - \tan {\theta \over 2}}}} \right]$$
= tan-1 $$\left[ {{{\tan {\pi \over 4} + \tan {\theta \over 2}} \over {1 - \tan {\pi \over 4} + \tan {\theta \over 2}}}} \right]$$
= tan-1 $$\left( {\tan \left( {{\pi \over 4} + {\theta \over 2}} \right)} \right)$$
= $${\pi \over 4} + {\theta \over 2}$$
= $${\pi \over 4} + {1 \over 2}$$ cos-1 x2
tan-1 $$\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right]$$
Let x2 = cos $$\theta $$
= tan-1 $$\left[ {{{\sqrt {1 + \cos \theta } + \sqrt {1 - \cos \theta } } \over {\sqrt {1 + \cos \theta } - \sqrt {1 - \cos \theta } }}} \right]$$
= tan-1 $$\left[ {{{\sqrt {2{{\cos }^2}{\theta \over 2}} + \sqrt {2{{\sin }^2}{\theta \over 2}} } \over {\sqrt {2{{\cos }^2}{\theta \over 2}} - \sqrt {2{{\sin }^2}{\theta \over 2}} }}} \right]$$
= tan-1 $$\left[ {{{\sqrt 2 \cos {\theta \over 2} + \sqrt 2 \sin {\theta \over 2}} \over {\sqrt 2 \cos {\theta \over 2} - \sqrt 2 \sin {\theta \over 2}}}} \right]$$
= tan-1 $$\left[ {{{\cos {\theta \over 2} + \sin {\theta \over 2}} \over {\cos {\theta \over 2} - \sin {\theta \over 2}}}} \right]$$
= tan-1 $$\left[ {{{1 + \tan {\theta \over 2}} \over {1 - \tan {\theta \over 2}}}} \right]$$
= tan-1 $$\left[ {{{\tan {\pi \over 4} + \tan {\theta \over 2}} \over {1 - \tan {\pi \over 4} + \tan {\theta \over 2}}}} \right]$$
= tan-1 $$\left( {\tan \left( {{\pi \over 4} + {\theta \over 2}} \right)} \right)$$
= $${\pi \over 4} + {\theta \over 2}$$
= $${\pi \over 4} + {1 \over 2}$$ cos-1 x2
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