In Fig. find tan P-cot R . | vedclass

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Question

Applying Pythagoras theorem for $	riangle PQR$, we obtain

$PR ^{2}= PQ ^{2}+ QR ^{2}$

$(13\, cm )^{2}=(12\, cm )^{2}+ QR ^{2}$

$169 \,cm ^{2

Vedclass · Accepted Answer

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Vedclass · Answer

Applying Pythagoras theorem for $	riangle PQR$, we obtain

$PR ^{2}= PQ ^{2}+ QR ^{2}$

$(13\, cm )^{2}=(12\, cm )^{2}+ QR ^{2}$

$169 \,cm ^{2}=144 \,cm ^{2}+ QR ^{2}$

$25\, cm ^{2}= QR ^{2}$

$QR =5\, cm$

$	an P=\frac{	ext { Side opposite to } \angle P }{	ext { Side adjacent to } \angle P }=\frac{ QR }{ PQ }$

$=\frac{5}{12}$

$\cot R=\frac{	ext { Side adjacent to } \angle R }{	ext { Side opposite to } \angle R }=\frac{ QR }{ PQ }$

$=\frac{5}{12}$

$	an P- \cot R =\frac{5}{12}-\frac{5}{12}=0$

In $Fig.$ find $\tan P-\cot R .$

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