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 #sinx=1/6# and #cosx gt 0# then #x# is acuite.

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 #