How to simplify #sin(sec^(-1)(x))# ? Geometry Right Triangles and Trig Sine, Cosine and Tangent Functions 1 Answer Guillaume L. Aug 2, 2018 #sin(sec^(-1)(x))=sqrt(x^2-1)/x# Explanation: #sin(sec^(-1)(x))=sin(cos^(-1)(1/x))# Let #y=cos^(-1)(1/x)# #x=1/cos(y)# #1/x^2=cos(y)^2=1-sin(y)^2# #1/x^2-1=-sin(y)^2# #(x^2-1)/x^2=sin(y)^2# #sqrt(x^2-1)/x=sin(y)=sin(sec^(-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 #tan(sec^(-1)(x))# ? How do you simplify #sec(tan^(-1)(x))# ? See all questions in Sine, Cosine and Tangent Functions Impact of this question 33623 views around the world You can reuse this answer Creative Commons License