How do you evaluate #sin(tan^-1(12/5))#?

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]

Related questions
See all questions in Basic Inverse Trigonometric Functions
Impact of this question
16348 views around the world
You can reuse this answer
Creative Commons License