Question

In right triangle $PQR$, $\cot P=\frac{5}{12}$. Find the value of $\tan P,\sin P\sec P$.

Answer 1

Given, $\cot P=\frac{5}{12}$
$\tan P=\frac{1}{\cot P}=\frac{12}{5}$
$\sec P=\sqrt{1+{\tan }^{2}P}=\sqrt{1+{\left(\frac{12}{5}\right)}^2}=\sqrt{\frac{169}{25}}=\frac{13}{5}$
$\sin P=\sqrt{1-{\cos }^{2}P}=\sqrt{1-{\left(\frac{5}{13}\right)}^2}=\sqrt{\frac{144}{169}}=\frac{12}{13}$