How do you solve and find the value of #tan(cos^-1(6/7))#?

2 Answers
Feb 1, 2017

0.60

Explanation:

Method 1: Use calculator
#cos ^-1 (6/7) --> arccos (6/7)#
Calculator gives:
cos x = 6/7 --> #arc x = 31^@#
#tan 31^@ = 0.60#
Method 2: Use right triangle OMA
with horizontal leg OA = 6
hypotenuse OM = 7
Vertical leg #AM = sqrt(49 - 36) = sqrt13#
Call t the angle < AOM:
#tan t = AM/OA = sqrt13/6 = 0.60#

Feb 1, 2017

#1/sqrt3#

Explanation:

Let #a = cos^(-1)(6/7) in Q_1, cos a = 6/7 and tan a > 0#

So the given expression

#tan a = sina/cos a=sqrt(1-cos^2a)/cos a=sqrt(1-36/49)/(6/7)=1/sqrt3#