How do you evaluate #sin(tan^-1(12/5))#? Trigonometry Inverse Trigonometric Functions Basic Inverse Trigonometric Functions 1 Answer Binayaka C. Jan 31, 2017 #sin (tan^-1(12/5)) =12/13# Explanation: #sin (tan^-1(12/5))#. Let #tan^-1(12/5) = theta :. tan theta = 12/5# Since #tan^-1# exists in 1st quadrant & 4th quadrant and positive in 1st quadrant, #theta# is in 1st quadrant . we know #tan theta =p/b= 12/5 :. p=12 ; b=5 ; h= sqrt(p^2+b^2)=sqrt(12^2+5^2)=13 ; sin theta = p/h=12/13#. [p=perpendicular ; b = base; h= hypotenuse] #:.sin (tan^-1(12/5)) =sin theta = 12/13# [Ans] 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 17379 views around the world You can reuse this answer Creative Commons License