Question

If cos(θ)= -2/3 and θ lies in quadrant II, find the value of cotθ

Answer 1

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Answer 2

`\text{Given that,}\\\\\cos \theta = - \frac 23\\\\\implies \cos^2 \theta = \frac 49\\\\\implies 1- \sin^2 \theta = \frac 49\\\\\implies \sin^2 \theta =1-\frac 49\\\\\implies \sin^2 \theta =\frac 59\\\\\\\text{Now,}\\\\\cot^2 \theta = (\cos^2 \theta)/(\sin^2 \theta) \\\\\\\implies \cot^2 \theta = (\tfrac 49)/(\tfrac 59)\\\\\\\implies \cot^2 \theta = \frac 49 * \frac 95\\\\\\\implies \cot^2 \theta = \frac 45\\\\\\`

`\implies \cot \theta = \pm√(\frac 45) = \pm \frac 2{\sqrt 5}\\\\\\\text{Since}~ \theta ~ \text{lies in the second quadrant,} \cot \theta ~\text{will be negative.}\\\\\text{Hence,}~ \cot \theta = - (2)/(\sqrt 5)`