If sin(2x - 45°)= cos x and angle x and (2x - 45°) in degrees) are acute angles, then the value of cot x is:

0 $\frac{1}{2}$ 1 $\frac{1}{\sqrt{2}}$

1

sin(2x - 45°)= cos x We know , sin A = cos B So , A + B = 90º ⇒ sin(2x - 45°)= cos x  2x - 45° + x = 90° 3x = 135° x = \(\frac{135° }{3}\)  = 45° Now , cotx = cot 45° = 1