In PQR- PQ - 12 cm- PR - 13 cm and - Q - 90- Find tan P - cot R-

Given that in PQR, PQ = 12 cm and PR = 13 cm. Now, from Pythagoras theorem, PQ^2 + QR^2 = PR^2 ⇒QR^2 = PR^2 PQ^2 ⇒QR^2 = 13^2 12^2 ⇒QR^2 = 169 144 = 25 ...

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Standard X

Mathematics

Trigonometric Ratios of a Right Triangle

In PQR, PQ = ...

Question

In $△$ PQR, PQ = 12 cm, PR = 13 cm and $∠ Q = 90^{\circ}$ . Find tan P - cot R.

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Solution

Given that in $△$ PQR, PQ = 12 cm and PR = 13 cm.

Now, from Pythagoras theorem,
${P Q}^{2} + {Q R}^{2} = {P R}^{2}$
$\Rightarrow {Q R}^{2} = {P R}^{2} - {P Q}^{2}$
$\Rightarrow {Q R}^{2} = 13^{2} - 12^{2}$
$\Rightarrow {Q R}^{2} = 169 - 144 = 25$
$\Rightarrow Q R = \sqrt{25} = 5 cm$

Now, $t a n P = \frac{opposite side}{adjacent side} = \frac{Q R}{P Q} = \frac{5}{12}$
$c o t R = \frac{adjacent side}{opposite side} = \frac{Q R}{P Q} = \frac{5}{12}$

$∴ t a n P - c o t R = \frac{5}{12} - \frac{5}{12} = 0$

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In △PQR, PQ = 12 cm, PR = 13 cm and ∠Q=90∘. Find tan P - cot R.

Given that in △ PQR, PQ = 12 cm and PR = 13 cm. Now, from Pythagoras theorem, PQ2+QR2=PR2 ⇒QR2=PR2−PQ2 ⇒QR2=132−122 ⇒QR2=169−144=25 ⇒QR=√25=5 cm Now, tan P=opposite sideadjacent side=QRPQ=512 cot R=adjacent sideopposite side=QRPQ=512 ∴tan P−cot R=512−512=0

In $△$ PQR, PQ = 12 cm, PR = 13 cm and $∠ Q = 90^{\circ}$ . Find tan P - cot R.