Question

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

Answer 1

$Given \: in \: triangle \:PQR , \angle Q = 90\degree$

$PR = 41 , \:and \: PQ - QR = 31$

$Let \: PQ = x \: and \: QR = y$

$PQ - QR = 31 \implies x - y = 31$

$\implies x = y + 31 \: --(1)$

/* By Pythagoras Theorem */

$PQ^{2} + QR^{2} = PR^{2}$

$\implies x^{2} + y^{2} = 41^{2}$

$\implies (y + 31)^{2} + y^{2} = 41^{2} \: [ From \: (1) ]$

$\implies y^{2} + 61y + 961 + y^{2} = 1681$

$\implies 2y^{2} + 62y + 961 - 1681= 0$

$\implies 2y^{2} + 62y - 720= 0$

/* Dividing each term by 2 , we get */

$\implies y^{2} + 31y - 360 = 0$

/* Splitting the middle term,we get */

$\implies y^{2} + 40y - 9y-360 =0$

$\implies y( y + 40 ) - 9( y + 40 ) = 0$

$\implies (y+40)(y-9) = 0$

$\implies y+40 = 0 \: Or \: y-9= 0$

$\implies y = -40 \: Or \: y = 9$

/* y should not be negative . */

$\therefore y = 9 \: --(2)$

$put \: y = 9 \: in \: equation \: (1) , we \:get$

$\implies x = 9 + 31$

$\implies x = 40 \: --(3)$

$Now , \red{ Value \:of \: Sec 2R - tan 2R }$

$= \frac{1}{Cos 2R} - \frac{sin 2R}{cos 2R}$

$= \frac{1 - sin 2R}{cos 2R}$

$= \frac{ 1 - 2 sin R cos R}{ cos^{2} R - sin^{2} R}$

$= \frac{ 1- 2 \times \frac{40}{41} \times \frac{9}{41} }{ \Big ( \frac{9}{41}\Big)^{2} -\Big( \frac{40}{41}\Big)^{2} }$

$= \frac{ \frac{1681 - 720}{1681}}{ \frac{81 - 1600}{1681} }$

$= \frac{ 961}{-1519}$

Therefore.,

$\red{ Value \:of \: Sec 2R - tan 2R }$

$\green { = \frac{- 961}{1519} }$

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