How do you rewrite #sec^2 (tan^{-1} x)# in terms of x? Trigonometry Inverse Trigonometric Functions Inverse Trigonometric Properties 1 Answer Wataru Oct 24, 2014 Let #theta =tan^{-1}x Leftrightarrow tan theta=x#. So, #sec^2(tan^{-1}x)=sec^2theta=1+tan^2theta=1+x^2#. I hope that this was helpful. Answer link Related questions How do you use the properties of inverse trigonometric functions to evaluate #tan(arcsin (0.31))#? What is #\sin ( sin^{-1} frac{sqrt{2}}{2})#? How do you find the exact value of #\cos(tan^{-1}sqrt{3})#? How do you evaluate #\sec^{-1} \sqrt{2} #? How do you find #cos( cot^{-1} sqrt{3} )# without a calculator? How do you use the inverse trigonometric properties to rewrite expressions in terms of x? How do you calculate #sin^-1(0.1)#? How do you solve the inverse trig function #cos^-1 (-sqrt2/2)#? How do you solve the inverse trig function #sin(sin^-1 (1/3))#? How do you solve the inverse trig function #arcsin (sin 5pi/6)#? See all questions in Inverse Trigonometric Properties Impact of this question 18194 views around the world You can reuse this answer Creative Commons License