Question

Question 2
In the figure, find tan P – cot R.

Answer 1

By applying Pythagoras theorem in $\Delta$ PQR, we get;

$P{R}^2=P{Q}^2+Q{R}^2$
$(13{)}^2=(12{)}^2+Q{R}^2$

$169=144+Q{R}^2$
$\Rightarrow Q{R}^2=25\Rightarrow$QR = 5 cm
Now,
$tanP=\frac{QR}{PQ}=\frac{5}{12}$
$cotR=\frac{QR}{PQ}=\frac{5}{12}$
According to the question,
$tanP-cotR=\frac{5}{12}-\frac{5}{12}=0$