How do you use fundamental identities to find the values of the trigonometric values given θ be an angle in quadrant III, such that cot θ = (√11)/(5)?

1 Answer

Answer:

The given #cot theta=sqrt(11)/5#,
the answers #tan theta=5/sqrt(11)#, #" "sin theta=-5/6#, #" "cos theta=-sqrt(11)/6#,#" "csc theta=-6/5#, #" "sec theta=(-6sqrt11)/11#

Explanation:

the given #cot theta# at the third quadrant x is negative and y is negative.

The formulas:
#tan theta=1/cot theta#
#csc theta=+-sqrt(cot^2 theta+1)#
#sin theta=1/csc theta=1/(+-sqrt(cot^2 theta+1))#
#cos theta=sin theta*cot theta=cot theta/(+-sqrt(cot^2 theta+1))#
#sec theta=1/cos theta=(+-sqrt(cot^2 theta+1))/cot theta#

Compute the values of the other remaining trigonometric functions

Start with #cot theta=(-sqrt(11))/(-5)#
#tan theta=1/cot theta=1/((-sqrt(11))/(-5))=5/sqrt(11)=(5sqrt11)/11#

#csc theta=+-sqrt(cot^2 theta+1)=+-sqrt(((-sqrt(11))/(-5))^2+1)=+-sqrt(11/25+1)=-6/5#

#sin theta=1/csc theta=1/(+-sqrt(cot^2 theta+1))=1/(+-sqrt(((-sqrt(11))/(-5))^2+1))=1/(+-sqrt((11/25+1)))=-5/6#

#cos theta=sin theta*cot theta=cot theta/(+-sqrt(cot^2 theta+1))((-sqrt(11))/(-5))/(+-sqrt(((-sqrt(11))/(-5))^2+1))=((-sqrt(11))/(-5))/(+-6/5)=-sqrt(11)/6#

#sec theta=1/cos theta=(+-sqrt(cot^2 theta+1))/cot theta=+-sqrt(((-sqrt(11))/(-5))^2+1)/((-sqrt(11))/(-5))=(-6/5)/((-sqrt(11))/(-5))=(-6)/sqrt(11)=(-6sqrt11)/11#

God bless....I hope the explanation si useful.