Question

Given: $\cos A=\frac{5}{13}$
Evaluate: (i) $\frac{\sin A-\cot A}{2\tan A}$ (ii) $\cot A+\frac{1}{\cos A}$

Answer 1

$\cos A=\frac{5}{13}=\frac{B}{H}\sin A=\frac{P}{H}=\frac{\sqrt{H^2-B^2}}{H}=\frac{\sqrt{{13}^2-5^2}}{13}=\frac{12}{3}\tan A=\frac{P}{B}=\frac{12}{5}\left(1\right)\frac{\sin A-\cot A}{2\tan A}=\frac{\frac{12}{13}-\frac{5}{12}}{2\times \frac{12}{5}}=\frac{\frac{144-65}{12\times 13}}{\frac{2\times 12}{5}}=\frac{79}{13}\times \frac{5}{2}=\frac{395}{26}\left(2\right)\cot A+\frac{1}{\cos A}=\frac{5}{12}+\frac{13}{5}=\frac{25+(12\times 13)}{60}=\frac{181}{60}$