Trigonometry Important Question #6

This CBSE class 10 Maths extra question is from the topic Trigonometry. Concept Covered: Calculating trigonometric values and using quadratic equations & Pythagoras theorem.

Question 6 : In triangle PQR, right angled at Q if PR = 41 units and PQ – QR = 31. Find sec²R – tan²R.

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Video Explanation

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Explanatory Answer | Trigonometry Extra Question 6

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

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

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Step 2: Calculate sec²R – tan²R

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)} × \text{(41 - 40)}}{\text{81}}) =\\frac{\text{81}}{\text{81}}) = 1

So, sec²R – tan²R = 1