Find cot x and sec x give that sin x= 1/6 and cos x >0?
1 Answer
May 14, 2018
# cot x = sqrt(35)#
# sec x = (6sqrt(35))/35 #
Explanation:
If we consider the graphs of sinx and cosx, at the same scale, over the range ][0,2pi]]
graph{sinx [-0.5, 6.3, -2, 2]}
graph{cosx [-0.5, 6.3, -2, 2]}
Then we can conclude that if
So then using:
# sin^2x + cos^2x -= 1 => cos^2x = 1 - sin^2x #
# :. cos^2x = 1 - (1/6)^2 = 35/36#
# => cos x = sqrt(35/36) = sqrt(35)/6#
So, can write
# cot x = cosx/sinx #
# \ \ \ \ \ \ \ = (sqrt(35)/6)/(1/6) #
# \ \ \ \ \ \ \ = sqrt(35)#
And:
# sec x = 1/cosx #
# \ \ \ \ \ \ \ = 1/(sqrt(35)/6) #
# \ \ \ \ \ \ \ = 6/sqrt(35) #
# \ \ \ \ \ \ \ = (6sqrt(35))/35 #