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8. Introduction to Trigonometry
medium
In $Fig.$ find $\tan P-\cot R .$
A
$1$
B
$5$
C
$0$
D
$12$
Solution
Applying Pythagoras theorem for $\triangle PQR$, we obtain
$PR ^{2}= PQ ^{2}+ QR ^{2}$
$(13\, cm )^{2}=(12\, cm )^{2}+ QR ^{2}$
$169 \,cm ^{2}=144 \,cm ^{2}+ QR ^{2}$
$25\, cm ^{2}= QR ^{2}$
$QR =5\, cm$
$\tan P=\frac{\text { Side opposite to } \angle P }{\text { Side adjacent to } \angle P }=\frac{ QR }{ PQ }$
$=\frac{5}{12}$
$\cot R=\frac{\text { Side adjacent to } \angle R }{\text { Side opposite to } \angle R }=\frac{ QR }{ PQ }$
$=\frac{5}{12}$
$\tan P- \cot R =\frac{5}{12}-\frac{5}{12}=0$
Standard 10
Mathematics
