Solution:
By using Pythagoras theorem in \[\vartriangle PQR\], we have
\[P{{R}^{2}}~=\text{ }P{{Q}^{2}}~+\text{ }Q{{R}^{2}}\]
Putting the length of given side PR and PQ in the above equation
\[{{13}^{2~}}=\text{ }{{12}^{2}}~+\text{ }Q{{R}^{2}}\]
\[Q{{R}^{2}}~=\text{ }{{13}^{2}}~\text{ }{{12}^{2}}\]
\[Q{{R}^{2}}~=\text{ }169\text{ }\text{ }144\]
\[Q{{R}^{2~}}=\text{ }25\]
\[QR\text{ }=~\surd 25\text{ }=\text{ }5\]
By definition,
tan P = Perpendicular side opposite to P/ Base side adjacent to angle P
\[tan\text{ }P\text{ }=\text{ }QR/PQ\]
\[tan\text{ }P\text{ }=\text{ }5/12~\ldots \ldots \ldots .\text{ }\left( 1 \right)\]
And,
cot R= Base/Perpendicular
\[cot\text{ }R=\text{ }QR/PQ\]
\[cot\text{ }R=\text{ }5/12~\ldots .\text{ }\left( 2 \right)\]
When comparing \[equation\text{ }\left( 1 \right)\text{ }and\text{ }\left( 2 \right)\], we can see that R.H.S of both the equation is equal.
Therefore, L.H.S of both equations should also be equal.
\[\therefore tan\text{ }P\text{ }=\text{ }cot\text{ }R\]
Yes, \[\mathbf{tan}\text{ }\mathbf{P}\text{ }=\text{ }\mathbf{cot}\text{ }\mathbf{R}\text{ }=~\mathbf{5}/\mathbf{12}\]