We're told that #sintheta# is negative; however, #cottheta=costheta/sintheta# is positive.
This tells us that both sine and cosine must be negative (the negatives cancel out to give a positive cotangent), so we're working in the fourth quadrant.
Then, #tantheta=sintheta/costheta# will also be positive.
Furthermore, since #sectheta=1/costheta, sectheta# will also be negative as a result of the negative cosine.
Keeping this in mind, we'll continue.
Recall the identity
#sin^2theta+cos^2theta=1#
#sintheta=(-7/8), sin^2theta=(-7/8)^2=49/64#
Thus,
#49/64+cos^2theta=64/64#
#cos^2theta=(64-49)/64#
#cos^2theta=15/64#
#costheta=+-sqrt(15/64)=+-sqrt15/8#
We want the negative result:
#costheta=-sqrt15/8#
Then,
#sectheta=1/costheta=1/(sqrt15/8)=-8/sqrt15#
Rationalizing the denominator (multiply by #sqrt15/sqrt15#), we get
#sectheta=-(8sqrt15)/15#
We can now find the tangent:
#tantheta=sintheta/costheta#
#tantheta=(cancel-7/8)/(cancel-(8sqrt15)/15)#
#tantheta=7/8*15/(8sqrt15)#
#tantheta=105/(64sqrt15)#
Rationalizing, we get
#tantheta=(105sqrt15)/(960)#
#tantheta=(7sqrt15)/(64)#
