Question
Download Solution PDFयदि 2sin(3x -15)° = 1 है, जहाँ 0° < (3x - 15) < 90° है, तो cos2(2x +15)° + cot2 (x + 15)° का मान ज्ञात कीजिए।
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया है:
2sin(3x -15)° = 1
गणना:
sin(3x - 15)° = 1/2
⇒ sin (3x - 15)° = sin 30°
⇒ (3x - 15)° = 30°
⇒ 3x = 30° + 15° = 45°
⇒ x = \(45\over3\) = 15°
तब,
cos2(2x +15)° + cot2 (x +15)°
cos2(2 × 15 +15)° + cot2 (15 +15)°
⇒ cos2(45)° + cot2(30)°
⇒ \(({1\over \sqrt2})^2\) + \((\sqrt3)^2\)
⇒ \(1\over2\) + 3 = \(7\over2\)
∴ cos2(2x +15)° + cot2 (x +15)° = \(7\over2\)
Last updated on Jun 13, 2025
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