Question

In triangle PQR,right angled at Q if PR=41 units and PQ-QR=31 find sec2R-tan2R​

Answer 1

Step-by-step explanation:

Qr not be negative so Qr =9

sec2= (41/9)2= 1681/81

tan2= (40/9)2= 1600/81

A/Q,

1581/81-1600/81= 81/81= 1 Answer

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Answer 2

The given question is In triangle PQR,right angled at Q if PR=41 units and PQ-QR=31

we have to find the value of sec2R-tan2R.

In a triangle PQR

angle Q is 90°, PR=41, PQ-QR=31.

By Pythagoras theorem

${hypotense}^{2} = {base}^{2} + {height}^{2} \\ {pr}^{2} = {pq}^{2} + {qr}^{2} \\ {pq}^{2} = {(31 + qr)}^{2} + {qr}^{2} \\ {41}^{2} = 961 + {qr}^{2} + 62qr + {qr}^{2} \\ 1681 = 961 + 2 {qr}^{2} + 62qr \\ 1681 - 961 = 2 {qr}^{2} + 62qr \\ 720 = 2 {qr}^{2} + 62qr \\ = 2 {qr}^{2} + 62qr - 720 \\ \div \: \: by \: 2 \: we \: get \\ = {qr}^{2} + 32qr - 360 \\ on \: factorising \: we \: get \\ {qr}^{2} + 40qr - 9qr - 360 \\ qr(qr + 40) - 9(qr + 40) \\ = (qr - 9)(qr + 40) \\ qr = 9 \\ qr = - 40$

QR cannot be negative so we take QR=9

sub the value of QR in PQ-QR=31

PQ=31+9=40.

let's find the value for

${sec}^{2} \\ {tan}^{2}$

sec is hypotenuse÷ adjacent.

tan is opposite÷adjacent so substituting in the formula we get.

$sec = \frac{41}{9} \\ tan = \frac{40}{9}$

on squaring and subtracting we get,

${sec}^{2} r - {tan}^{2} r = { ( \frac{41}{9} )}^{2} - { (\frac{40}{9}) }^{2} \\ = \frac{1681}{81} - \frac{1600}{81} \\ \frac{1681 - 1600}{81} \\ = \frac{81}{81} = 1$

Therefore the final value is 1

${sec}^{2} r - {tan}^{2} r = 1$

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