Question

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

Answer 1

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`

Answer 2

`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`