We have to evaluate \(\sec (tan^{-1}x)\)
Solution
\(\sec (tan^{-1}x)\)Let us assume that
\(y=(tan^{-1}x)\)
x = tan y
x = sin y / cos y
\(x^{2}=\frac{(sin y)^{2}}{\(\cos y)^{2}}\) \(x^{2}+ 1=\frac{\cos ^{2}y+ \sin ^{2}y}{\cos ^{2}y}\) \(x^{2}+ 1=\sec ^{2}y\) \(\sqrt{x^{2}+ 1}=\sec ^{2}y\) \(\sqrt{x^{2}+ 1}=\sec (\tan ^{-1}x)\) \(\sec (\tan ^{-1}x)=\sqrt{x^{2}+ 1}\)Answer
\(\sec (\tan ^{-1}x)=\sqrt{x^{2}+ 1}\)
