In the given figure,find tan P - cot R. | vedclass

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(C) In $	riangle PQR$,$\angle Q = 90^{\circ}$,$PQ = 12 \, 	ext{cm}$,and $PR = 13 \, 	ext{cm}$.Applying the Pythagoras theorem:$PR^2 = PQ^2 + QR^2$$

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(C) In $	riangle PQR$,$\angle Q = 90^{\circ}$,$PQ = 12 \, 	ext{cm}$,and $PR = 13 \, 	ext{cm}$.Applying the Pythagoras theorem:$PR^2 = PQ^2 + QR^2$$(13)^2 = (12)^2 + QR^2$$169 = 144 + QR^2$$QR^2 = 169 - 144 = 25$$QR = 5 \, 	ext{cm}$.Now,$	an P = \frac{	ext{side opposite to } \angle P}{	ext{side adjacent to } \angle P} = \frac{QR}{PQ} = \frac{5}{12}$.And,$\cot R = \frac{	ext{side adjacent to } \angle R}{	ext{side opposite to } \angle R} = \frac{QR}{PQ} = \frac{5}{12}$.Therefore,$	an P - \cot R = \frac{5}{12} - \frac{5}{12} = 0$.

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

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