Answer:
tan(θ)=(√15/7)
What will be the sign of tangent value in 3rd quadrant?
The sign of tangent value in 3rd quadrant will be positive.
cos(θ)=-7/8
(sin(θ))^2
=1-(cos(θ))^2
=1-(7/8)^2
=1-(49/64)
=(64-49)/64
=15/64
sin(θ)=[tex]-\sqrt{15/64}[/tex]
[tex]=-\sqrt{15}/8[/tex]
[since in quadrant III value of sin(θ) will be negative]
so, tan(θ)=sin(θ)/cos(θ)=[tex]\frac{\frac{-\sqrt{15}}{8} }{-\frac{7}{8} }[/tex]=[tex]\frac{\sqrt{15} }{7}[/tex]
tan(θ)=(√15/7)
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