In the triangle shown- the value of tan P - R is

By applying Pythagoras theorem in Δ PQR, we get PR^2 = PQ^2 + QR^2 (13)^2 = (12)^2+QR^2 169 = 144 + QR^2 ⇒ QR^2 = 25⇒QR = 5 cm Now, tan P = QR/PQ = 5/12 cot ...

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Standard IX

Mathematics

Trigonometric Ratios

In the triang...

Question

In the triangle shown, the value of tan P – cot R is ___.

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Solution

By applying Pythagoras theorem in $Δ$ PQR, we get

$P R^{2} = P Q^{2} + Q R^{2} (13)^{2} = (12)^{2} + Q R^{2} 169 = 144 + Q R^{2}$
$\Rightarrow Q R^{2} = 25 \Rightarrow$ QR = 5 cm
Now,
$t a n P = \frac{Q R}{P Q} = \frac{5}{12}$
$c o t R = \frac{Q R}{P Q} = \frac{5}{12}$
According to question,
$t a n P - c o t R = \frac{5}{12} - \frac{5}{12} = 0$

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In the triangle shown, the value of tan P – cot R is ___.

By applying Pythagoras theorem in Δ PQR, we get PR2=PQ2+QR2(13)2=(12)2+QR2169=144+QR2 ⇒QR2=25⇒QR = 5 cm Now, tan P=QRPQ=512 cot R=QRPQ=512 According to question, tan P − cot R=512−512=0