if secA=5/3 verify that 4cos3A-3cosA/3sinA-4sin3A
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13/46
Step-by-step explanation:
cos A = 3/5 (reciprocal of sec A)
sin A = √cos²A + 1 = √(3/5)² + 1 = √9/25 + 1 = √36/25 = 6/5
Putting these values in the given equation ,
4cos3A-3cosA/3sinA-4sin3A
= 4×(3/5)³ - 3×(3/5) / 3×(6/5) - 4×(6/5)³
= 4×(27/125) - 9/5 / 18/5 - 4×(216/125)
= 108/125 - 9/5 / 18/5 - 864/125
= (108-225)/125 / (450-864)/125
= -117/125 / -414/125
= -117/125 × 125/-414
By cancelling the common factors like 125 , '-' and 9 ,
= 13/46
Hope this will help you.
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