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)#