Question

Proove: $\cot \theta =\sqrt{7}$
when $\frac{\cos \theta }{\sin \theta }=\sqrt{7}$

Answer 1

Given:$\frac{\cos \theta }{\sin \theta }=\sqrt{7}$ ; Prove:$\cot \theta =\sqrt{7}$

$\cos \theta =\frac{adjacent}{hypotenuse}$

$\sin \theta =\frac{opposite}{hypotenuse}$

$\frac{\cos \theta }{\sin \theta }=\frac{\sqrt{7}}{1}=\frac{adjacent}{opposite}$

By pythagoras theorem

$Hypothenu{s}^2=opposit{e}^2+adjacent2$

$H=\sqrt{7+1}=\sqrt{8}$

$\cot \theta =\frac{adjustment}{opposite}=\frac{\sqrt{7}}{1}=\sqrt{7}$