Evaluate the value of the expression: 3(sin θ − cos θ)⁴ + 6(sin θ + cos θ)² + 4(sin⁶θ + cos⁶θ).
व्यंजक (Expression) का मान ज्ञात कीजिए: 3(sin θ − cos θ)⁴ + 6(sin θ + cos θ)² + 4(sin⁶θ + cos⁶θ)
Answer और ExplanationAnswer and explanation
Answer: (C) 13
मान लें t = sin θ cos θ
(sin θ − cos θ)² = 1 − 2t ⇒ (sin θ − cos θ)⁴ = (1 − 2t)² = 1 − 4t + 4t²
(sin θ + cos θ)² = 1 + 2t
sin⁶θ + cos⁶θ = (sin²θ + cos²θ)³ − 3 sin²θ cos²θ(sin²θ + cos²θ) = 1 − 3t²
व्यंजक (Expression) = 3(1 − 4t + 4t²) + 6(1 + 2t) + 4(1 − 3t²)
= 3 − 12t + 12t² + 6 + 12t + 4 − 12t² = 13
✓
मान 13 है।
Let t = sin θ cos θ.
(sin θ − cos θ)² = 1 − 2t ⇒ (sin θ − cos θ)⁴ = (1 − 2t)² = 1 − 4t + 4t²
(sin θ + cos θ)² = 1 + 2t
sin⁶θ + cos⁶θ = (sin²θ + cos²θ)³ − 3 sin²θ cos²θ(sin²θ + cos²θ) = 1 − 3t²
Expression = 3(1 − 4t + 4t²) + 6(1 + 2t) + 4(1 − 3t²)
= 3 − 12t + 12t² + 6 + 12t + 4 − 12t² = 13
✓
The value is 13.
SMART APPROACH
उत्तर θ पर निर्भर नहीं है, इसलिए θ = 0° रखें: 3(0 − 1)⁴ + 6(0 + 1)² + 4(0 + 1) = 3 + 6 + 4 = 13।
The answer does not depend on θ, so put θ = 0°: 3(0 − 1)⁴ + 6(0 + 1)² + 4(0 + 1) = 3 + 6 + 4 = 13.
Source: Selection Post Phase XIV (Graduate) · 19 Sep 2026 · Shift 1। Solution: Prashant Sir, Vijayastra।Source: Selection Post Phase XIV (Graduate) · 19 Sep 2026 · Shift 1. Solution by Prashant Sir, from Vijayastra.