Given that $\theta$ is an acute angle with
$\sin (\theta)=\frac{17}{41}.$
find the exact value of $\cot (\theta)$
$\endgroup$ 43 Answers
$\begingroup$Hint: $\cos^2\theta+\sin^2\theta=1$ . Use this to get $\cos\theta$.
Then $\cot\theta =\frac{\cos\theta}{\sin\theta}$...
$\endgroup$ $\begingroup$$$\cot \theta=\frac{\cos \theta}{\sin\theta}=\frac{\frac{4\sqrt{87}}{41}}{\frac{17}{41}}=\frac{4\sqrt{87}}{17}$$
$\endgroup$ 4 $\begingroup$$$sin \theta=17/41 \implies $$
$$ cos \theta = \sqrt {1-sin^2 \theta }= 4\sqrt {87}/41\implies $$
$$cot\theta = \frac {cos \theta}{sin \theta} = 4\sqrt {87}/17 $$
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