Question

In the figure above, find $\tan P-\cot R$

• A. $5$
• B. $\frac{1}{13}$
• C. $0$
• D. $Noneofthese$
  

Answer 1

Answer: (C)

$0$

In the figure, by Pythagoras Theorem,
$Q{R}^2=P{R}^2-P{Q}^2$

$=(13{)}^2-(12{)}^2$

$=25$
$\Rightarrow QR=\sqrt{25}=5$ cm
In $△PQR$ right angled at $Q$,

$QR=5$ cm is side opposite to the angle $P$ and $PQ=12$ cm is side adjacent to the angle $P$.
Therefore, $\tan P=\frac{QR}{PQ}=\frac{5}{12}$
Now, $QR=5$ cm is side adjacent to the angle $R$ and $PQ=12$ cm is side opposite to the angle $R$.
Therefore, $\cot R=\frac{QR}{PQ}=\frac{5}{12}$.
Hence, $\tan P-\cot R=\frac{5}{12}-\frac{5}{12}=0$