Question

( begin{aligned}left(4 cos ^{2} 9^{circ}-3right)left(4 cos ^{2} 27^{circ}-3right) &left(4 cos ^{3} 27^{circ}-3 cos 27^{circ}right) =& frac{left(4 cos ^{3} 9^{circ}-3 cos 9^{circ}right)left(90^{circ}-9^{circ}right)}{cos 9^{circ}} =& frac{cos 27^{circ} cos 81^{circ}}{cos 9^{circ} cos 27^{circ}}=frac{cos 27^{circ}}{frac{sin 9^{circ}}{cos 9^{circ}}=tan 9^{circ}} Rightarrow mid K=9 end{aligned} )

# 31. If (4cos29° - 3)(4cos227° - 3) = tankº, then K is equal to

Solution