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In the figure, find $tan\ P$ and $cot\ R$. Is $tan\ P = cot\ R$?"
Given:
$PQ = 12\ cm, PR = 13\ cm$.
To do:
We have to find $tan\ P$ and $cot\ R$ and check whether $tan\ P = cot\ R$.
Solution:
We know that,
In a right-angled triangle $PQR$ with right angle at $B$,
By Pythagoras theorem,
$PR^2=PQ^2+QR^2$
By trigonometric ratios definitions,
$tan\ P=\frac{Opposite}{Adjacent}=\frac{QR}{PQ}$
$cot\ R=\frac{Adjacent}{Opposite}=\frac{QR}{PQ}$
Here,
$PR^2=PQ^2+QR^2$
$\Rightarrow (13)^2=(12)^2+QR^2$
$\Rightarrow QR^2=169-144$
$\Rightarrow QR=\sqrt{25}=5$
Therefore,
$tan\ P=\frac{QR}{PQ}=\frac{5}{12}$
$cot\ R=\frac{QR}{PQ}=\frac{5}{12}$
Hence, $tan\ P = cot\ R$.
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