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JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

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The sides of a △ABC\triangle A B C are in AP\mathrm{AP}, if the angles AA and CC are the greatest and the smallest angles respectively, then 4(1−cos⁡A)(1−cos⁡C)4(1-\cos A)(1-\cos C) is equal to
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Correct option: B
4cos⁡2A+C24 \cos ^2 \frac{A+C}{2}
Option B (4cos⁡2A+C24 \cos ^2 \frac{{A+C}}{{2}}) is correct because by using the properties of arithmetic progressions and trigonometric identities, we can show that 4(1−cos⁡A)(1−cos⁡C)4(1-\cos A)(1-\cos C) simplifies to 4cos⁡2A+C24 \cos ^2 \frac{{A+C}}{{2}}.

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