Question

Given sec(theta)= 5... find cot(90-theta)

Answer 1

$\sec(\theta)=\frac{1}{\cos(\theta)} \\ \\\cos(\theta)=\frac{1}{\sec(\theta)}=\frac{1}{5} \\ \\ \sin^2\theta+\cos^2\theta=1 \\ \\\sin\theta= \sqrt{1-\cos^2\theta}= \sqrt{1-( \frac{1}{5})^2 } = \sqrt{1- \frac{1}{25} } = \sqrt{ \frac{25}{25} -\frac{1}{25} }=\frac { \sqrt{24}}{5} \\ \\ \cot(90^o-\theta)=\tan{\theta}= \frac{\sin\theta}{\cos\theta} = \frac{\frac { \sqrt{24}}{5} }{ \frac{1}{5} } = \sqrt{24} = \sqrt{4\times6}=2 \sqrt{6}$