Question

If cos θ = -8/17, and 180° < θ < 270°, what is tan θ?

Answer 1

`\large\boxed{\tan\theta=(8)/(15)}`

Explanation:

`180^o<\theta<270^o\to\bold{Quadrant\ III}-\text{look at the picture}\\\\\tan\theta>0.\\\\\text{We have}\ \cos\theta=-(8)/(17).\\\\\tan\theta=(\sin\theta)/(\cos\theta)\\\\\text{Use}\ \sin^2\theta+\cos^2\theta=1:\\\\\sin^2\theta+\left(-(8)/(17)\right)^2=1\\\\\sin^2\theta+(64)/(289)=1\qquad\text{subtract}\ (64)/(289)\ \text{from both sides}\\\\\sin^2\theta=(289-64)/(289)\\\\\sin^2\theta=(225)/(289)\to\sin\theta=\pm\sqrt{(225)/(289)}`

`\sin\theta=\pm(√(225))/(√(289))\to\sin\theta=\pm(15)/(17)\\\\\bold{Quadrant\ III}\to\sin\theta<0\to\sin\theta=-(15)/(17)\\\\\text{Substitute to the formula of a tangent}:\\\\\tan\theta=(-(8)/(17))/(-(15)/(17))=(8)/(17\!\!\!\!\!\diagup)\cdot(17\!\!\!\!\!\diagup)/(15)=(8)/(15)`