How do you evaluate #tan (Arc cos (sqrt2/2) )#?

1 Answer
A. S. Adikesavan
Jun 8, 2016

#+-1#. It is 1, if #a = arc cos (sqrt 2/2)# is restricted to be in the 1st quadrant...

Explanation:

Let #a=arc cos (sqrt 2/2)=arc cos (1/sqrt 2)#. As cos (+-a)= cos a, a is in the 1st quadrant or in the 4th. Accordingly, #sin a = +-1/sqrt 2 and, therefore, tan a =+-1#.

if #a = arc cos (sqrt 2/2)# is restricted to be in the 1st quadrant. the given expression tan a = 1...

Related questions
See all questions in Basic Inverse Trigonometric Functions
Impact of this question
1098 views around the world
You can reuse this answer
Creative Commons License