How do you calculate # cos(tan^-1(3/4))#? Trigonometry Inverse Trigonometric Functions Basic Inverse Trigonometric Functions 2 Answers Binayaka C. May 27, 2018 # cos ( tan^-1 (3/4))= 0.8# Explanation: # cos ( tan^-1 (3/4))= ?# Let # tan^-1 (3/4)= theta# #:. tan theta = 3/4 =P/B, P and B # are perpendicular and base of right triangle , then #H^2= P^2+B^2= 3^2+4^2=25# #:.H=5 ; :. cos theta = B/H=4/5 =0.8# # cos ( tan^-1 (3/4))= cos theta= 0.8# # :. cos ( tan^-1 (3/4))= 0.8# [Ans] Answer link MattyMatty May 27, 2018 #4/5# Explanation: #tan(tan^-1(3/4)) = 3/4# #"Name "y = tan^-1(3/4)# #"Then we have"# #tan(y) = 3/4# #"Now use "sec²(x) = 1 + tan²(x)# #=> sec²(y) = 1 + tan²(y) = 1 + 9/16 = 25/16# #=> sec(y) = 1/cos(y) = pm 5/4# #=> cos(y) = pm 4/5# #=> cos(tan^-1(3/4)) = pm 4/5# #"We have to take the solution with + sign as"# #-pi/2 <= arctan(x) <= pi/2# #"and"# #cos(x) > 0, if -pi/2<=x<=pi/2# #=> cos(tan^-1(3/4)) = 4/5# #"Note that we could also have used"# #tan(y)=sin(y)/cos(y)# #"and"# #sin^2(y) + cos^2(y) = 1# #tan(y)=sin(y)/cos(y) = 3/4# #=> pm sqrt(1-cos^2(y))/cos(y) = 3/4# #=> 1-cos^2(y) = ((3/4) cos(y))^2# #=> (1+9/16) cos^2(y) = 1# #=> cos^2(y) = 16/25# #=> cos(y) = 4/5# 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 56560 views around the world You can reuse this answer Creative Commons License