1. In triangle PQR, right angled at Q if PR = 41 units and PQ – QR = 31 find sec 2 R – tan 2 R.
Answers
Answer:
Step-by-step explanation:
PQR with right angled at Q
PR = 41 units
PQ - QR = 31
PQ - QR = 31 → PQ = 31 + QR
Class 10 Trigonometry Q6 Triangle
Step 1: Compute QR and PQ
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
Factorize 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 + 21 = 40
Step 2: Calculate sec2R – tan2R
sec R = hypotenuseside adjacent to ∠R =PRQR= 419
tan R = side opposite to ∠Rside adjacent to ∠R = 409
sec2R – tan2R =41292–40292
(41 + 40)×(41 - 40)81 =8181 = 1
So, sec2R – tan2R = 1
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
Factorize 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 + 21 = 40
sec²R = (H/B)² = (41/9)² = 1681/81
tan²R = (P/B)² = (40/9)² = 1600/81
Now
sec²R -tan² R
= 1681/81 - 1600/81
= (1681-1600)
81
= 81
81
= 1
So, sec²R – tan²R = 1
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