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 sec^{2}R – tan^{2}R.

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PQR with right angled at Q

PR = 41 units

PQ - QR = 31

PQ - QR = 31 → PQ = 31 + QR

**Using Pythagoras Theorem, **PR^{2} = PQ^{2} + QR^{2}

41^{2} = (31 + QR)^{2} + QR^{2}

1681 = 961 + 62QR + QR^{2} + QR^{2}

2QR^{2} + 62QR – 720 = 0

Divide the equation by 2: QR^{2} + 31QR – 360 = 0

**Factorize** QR^{2} + 31QR – 360 = 0 QR^{2} + 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

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}})

sec^{2}R – tan^{2}R =\\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, ** sec ^{2}R – tan^{2}R = 1 **

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CBSE Class 10 Maths - 2021

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