Let PQR be a right angled triangle, right-angled at R. Let PQ = 29 cm, QR = 21 cm and ∠Q = θ. Find the value of cos2 θ – sin2 θ.

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SSC CGL 2024 Tier-II Official Paper-I (Held On: 20 Jan, 2025)
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  1. \(\frac{41}{841}\)
  2. \(\frac{841}{41}\)
  3. \(\frac{840}{40}\)
  4. \(\frac{40}{840}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{41}{841}\)
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Detailed Solution

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Given:

Let PQR be a right-angled triangle, right-angled at R.

PQ = 29 cm, QR = 21 cm, and ∠Q = θ.

Formula used:

cos² θ - sin² θ = cos 2θ

Calculations:

Task Id 1160 Daman (3)

In Triangle PQR using the Pythagorean theorem:

⇒ PR² = PQ² - QR²

⇒ PR² = 29² - 21²

⇒ PR² = 841 - 441

⇒ PR = 20 cm

cos θ = QR / PQ = 21 / 29

sin θ = PR / PQ = 20 / 29

Next, we use the identity for cos² θ - sin² θ:

⇒ cos² θ - sin² θ = (cos θ + sin θ) × (cos θ - sin θ)

⇒ cos² θ - sin² θ = (21/29)² - (20/29)²

⇒ cos² θ - sin² θ = (441/841) - (400/841)

⇒ cos² θ - sin² θ = (441 - 400) / 841 = 41 / 841

∴ The correct answer is option (1).

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