What is the value of (32 cos6 x – 48 cos4 x + 18 cos2 x – 1)/[4 sin x cos x sin (60 – x) cos (60 – x) sin (60 + x) cos (60 + x)]?
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Correct answer: C
Explanation
The correct answer is option 3 i.e. 8cot 6x
(32 cos6 x – 48 cos4 x + 18 cos2 x – 1)/[4 sin x cos x sin (60 – x) cos (60 – x) sin (60 + x) cos (60 + x)]
In numerator by adding and subtracting 4,
= 32 cos6 x – 4 - 48 cos4 x + 18 cos2 x – 1 + 4
= 32 cos6 x – 4 - 48 cos4 x + 24 cos2 x - 6 cos2 x + 3
= 4(8cos2 x - 1 - 12cos4 x + 6cos2 x) - 3(2cos2 x - 1)
= 4cos3 2x - 3cos2 x
= cos3(2x) = cos 6x
So,
cos 6x/[4 sin x cos x sin (60 – x) cos (60 – x) sin (60 + x) cos (60 + x)]
= cos 6x/[4 × 1/4 sin3x × 1/4 cos 3x]
= cos 6x/[1/8 × 2 sin3x × cos 3x]
= 8 cos 6x/sin 6x
= 8 cot 6x