Question

In Fig., find $\tan P-\cot R$

Answer 1

In $△PQR$

$P{R}^2=P{Q}^2+Q{R}^2$ ....(Pythagoras theorem)

${13}^2={12}^2+Q{R}^2$

$Q{R}^2=\sqrt{169-144}=\sqrt{25}=5$ cm

$\tan P=\frac{QR}{PQ}=\frac{5}{12},\cot R=\frac{QR}{PQ}=\frac{5}{12}$

$\therefore \tan P-\cot R=\frac{5}{12}-\frac{5}{12}=0$