What is the value of 32 ^6x - 48 ^4x + 18 ^2x - 14 x\, x\, ( 60 - x ) ( 60 - x ) ( 60 + x ) ( 60 + x )?

Option C. 8cot6x. Answer & Solution Answer: Option C Solution: & 32 ^6x - 48 ^4x + 18 ^2x - 14 x\, × x × \, ( 60 - x ) × ( 60 - x ) × ( 60 + x ) × ( 60 + x ) & Adding + 4\, and - 4, & 32 ^6x - 4 - 48 ^4x + 18 ^2x + 3 & 32 ^6x - 4 - 48 ^4x + 24 ^2x - 6 ^2x + 3 & 32 ^6x - 4 - 48 ^4x + 24 ^2x - 3 2x…

What is the value of 32 ^6x - 48 ^4x + 18 ^2x - 14 x\, x\, ( 60 - x )…

Answer: Option C

Solution:

\frac{32 cos ^{6} x - 48 cos ^{4} x + 18 cos ^{2} x - 1}{4 sin x \times cos x \times sin ( 60 - x ) \times cos ( 60 - x ) \times sin ( 60 + x ) \times cos ( 60 + x )} Adding + 4 and - 4, \Rightarrow 32 cos^{6} x - 4 - 48 cos^{4} x + 18 cos^{2} x + 3 \Rightarrow 32 cos^{6} x - 4 - 48 cos^{4} x + 24 cos^{2} x - 6 cos^{2} x + 3 \Rightarrow 32 cos^{6} x - 4 - 48 cos^{4} x + 24 cos^{2} x - 3 cos 2 x \Rightarrow 4 (2 cos^{2} x - 1)^{3} - 3 cos 2 x [∴ cos 2 x = 2 cos^{2} x - 1] \Rightarrow 4 cos^{3} 2 x - 3 cos 2 x \Rightarrow cos 6 x

∵ cos 3 x \to 4 cos^{3} x - 3 cos x and ∵ cos 6 x \to 4 cos^{3} 2 x - 3 cos 2 x

Now, 4 sin x . cos x . sin (60 - x) . cos (60 - x) . sin (60 + x) . cos (60 + x) [∴ sin x . sin (60 - x) . sin (60 + x) = \frac{1}{4} sin 3 x] [∴ cos x . cos (60 - x) . cos (60 + x) = \frac{1}{4} cos 3 x] = \frac{cos 6 x}{\frac{1}{4} sin 3 x . cos 3 x} = \frac{2}{2} \times \frac{4 cos 6 x}{sin 3 x . cos 3 x} = \frac{8 cos 6 x}{2 sin 3 x . cos 3 x} = \frac{8 cos 6 x}{sin 6 x} = 8 cot 6 x