How do you simplify #sec(tan^(-1)(x))# ? Geometry Right Triangles and Trig Sine, Cosine and Tangent Functions 1 Answer Guillaume L. Aug 8, 2018 #sec(tan^(-1)(x))=sqrt(x^2+1)# Explanation: #sec(tan^(-1)(x))# let #y=tan^(-1)(x)# #x=tan(y)# #x=sin(y)/cos(y)# #x^2=sin(y)^2/cos(y)^2# #x^2+1=(cancel(cos(y)^2+sin(y)^2)^(=1))/cos(y)^2# #x^2+1=sec(y)^2# #sqrt(x^2+1)=sec(y)=sec(tan^(-1)(x))# \0/ Here's our answer ! Answer link Related questions Why is the cosine of an obtuse angle negative? How would I solve #cos x + cos 2x = 0#? Please show steps. If A is an acute angle and sin A = .8406, what is angle A round to the nearest tenth of a degree? Question #a69e0 If #sin B = -12/13# then what is #cos 2B# ? An airplane is at height of #10000# feet. At what angle (rounded to whole degree), it must... Question #02b52 How to simplify #sin(sec^(-1)(x))# ? How to simplify #tan(sec^(-1)(x))# ? See all questions in Sine, Cosine and Tangent Functions Impact of this question 36797 views around the world You can reuse this answer Creative Commons License