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

PQR with right angled at Q
PR = 41 units
PQ - QR = 31
PQ - QR = 31 

PQ = 31 + QR

Using Pythagoras Theorem, PR² = PQ² + QR²
41² = (31 + QR)² + QR²
1681 = 961 + 62QR + QR² + QR²
2QR² + 62QR – 720 = 0
Divide the equation by 2: QR² + 31QR – 360 = 0

By Factorizing, we get :

QR² + 31QR – 360 = 0

 QR² + 40 QR – 9QR – 360 = 0
QR(QR + 40) – 9(QR + 40) = 0
(QR + 40) (QR – 9) = 0
QR = - 40 or QR = 9
QR cannot be negative ∴ QR = 9
PQ = QR + 31 = 9 + 31 = 40
 

Now, 

sec R = $\frac{\text{hypotenuse}}{\text{side adjacent to ∠R}}$ = $\frac{\text{PR}}{\text{QR}}$= $\frac{\text{41}}{\text{9}}$

tan R = $\frac{\text{side opposite to ∠R}}{\text{side adjacent to ∠R}}$ = $\frac{\text{40}}{\text{9}}$

sec²R – tan²R = $\frac{{41}^2}{9^2}$– $\frac{{40}^2}{9^2}$

                    = $\frac{\text{(41 + 40)}\times \text{(41 - 40)}}{\text{81}}$ = $\frac{\text{81}}{\text{81}}$  = 1

So, sec²R – tan²R = 1