If A is an angle in QIV such that sin A = − 3 5 and B is an angle in QIV such that sin B = − 1 3 use identities to find COS A COS B TAN(A+B) TAN(A-B) ?

We are given:

#sinA=-3/5#

Squaring both sides:

#sin^2A=9/25#

Identity:

#color(red)bb(sin^2x+cos^2x=1)#

i.e.

#sin^2x=1-cos^2x#

Substituting this:

#1-cos^2A=9/25#

#cos^2A=16/25#

Taking roots:

#cosA=+-(sqrt(16))/sqrt(25)=4/5#

Since we are in the IV quadrant we expect the cosine to be positive:

#color(blue)(cosA=4/5#

For #cosB# we use the same idea:

#sinB=-1/3#

#sin^2B=1/9#

#1-cos^2B=1/9#

#cos^2B=8/9#

Taking roots:

#cosB=+-(2sqrt(2))/3#

Since we are in the IV quadrant we expect the cosine to be positive:

#color(blue)(cosB=(2sqrt(2))/3)#

Using identity:

#color(red)bb(tanx=sinx/cosx)#

Find:

#tanA and tanB#

#tanA=sinA/cosA=(-3/5)/(4/5)=-3/4#

#tanB=sinB/cosB=(-1/3)/((2sqrt(2))/3)=-1/(2sqrt(2))=-sqrt(2)/4#

Identities:

#color(red)bb(tan(A+B)=(tanA+tanB)/(1-tanAtanB))#

#color(red)bb(tan(A-B)=(tanA-tanB)/(1+tanAtanB))#

#:.#

#tan(A+B)=((-3/4)+(-sqrt(2)/4))/(1-(-3/4)(-sqrt(2)/4))=((-3-sqrt(2))/4)/((16-3sqrt(2))/16)->#

#=(-12-4sqrt(2))/(16-3sqrt(2))=color(blue)(-((12+4sqrt(2)))/(16-3sqrt(2)))#

#tan(A-B)=((-3/4)-(-sqrt(2)/4))/(1+(-3/4)(-sqrt(2)/4))=((-3+sqrt(2))/4)/((16+3sqrt(2))/16)->#

#=(-12+4sqrt(2))/(16+3sqrt(2))=color(blue)(-((12-4sqrt(2)))/(16+3sqrt(2))#

.....................................................................................................................................

#color(blue)(cosA=4/5#

#color(blue)(cosB=(2sqrt(2))/3)#

#color(blue)(tan(A+B)=-((12+4sqrt(2)))/(16-3sqrt(2)))#

#color(blue)(tan(A-B)=-((12-4sqrt(2)))/(16+3sqrt(2))#