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Sec[tan^-1(y/2)], find its value

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Sec[tan^-1(y/2)], find its value
  • 2 answers

Sia ? 6 years, 2 months ago

Let {tex}{\tan ^{ - 1}}\frac{y}{2} = \theta{/tex}, where {tex}\theta \in \left( { - \frac{\pi }{2},\;\frac{\pi }{2}} \right){/tex}. So, {tex}\tan \theta = \frac{y}{2}{/tex}, which gives {tex}\sec \theta = \frac{{\sqrt {4 + {y^2}} }}{2}{/tex}.

Therefore, {tex}\sec \left( {{{\tan }^{ - 1}}\frac{y}{2}} \right) = \sec \theta = \frac{{\sqrt {4 + {y^2}} }}{2}{/tex}.

Miss Mor 6 years, 2 months ago

Value is under the root 4+y^2 by 2
http://mycbseguide.com/examin8/

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