How do you calculate #tan(arccos(5/13))#? Trigonometry Inverse Trigonometric Functions Basic Inverse Trigonometric Functions 1 Answer Massimiliano Feb 6, 2015 The answer is: #12/5#. If you set #alpha=arccos(5/13)rArrcosalpha=5/13# than we have to calculate: #tanalpha=sinalpha/cosalpha#. #cosalpha=5/13rArrsinalpha=sqrt(1-cos^2alpha)=sqrt(1-25/169)=# #=sqrt((169-25)/169)=sqrt(144/169)=12/13#. So: #tanalpha=(12/13)/(5/13)=12/13*13/5=12/5#. Answer link Related questions What are the Basic Inverse Trigonometric Functions? How do you use inverse trig functions to find angles? How do you use inverse trigonometric functions to find the solutions of the equation that are in... How do you use inverse trig functions to solve equations? How do you evalute #sin^-1 (-sqrt(3)/2)#? How do you evalute #tan^-1 (-sqrt(3))#? How do you find the inverse of #f(x) = \frac{1}{x-5}# algebraically? How do you find the inverse of #f(x) = 5 sin^{-1}( frac{2}{x-3} )#? What is tan(arctan 10)? How do you find the #arcsin(sin((7pi)/6))#? See all questions in Basic Inverse Trigonometric Functions Impact of this question 14254 views around the world You can reuse this answer Creative Commons License