If \cos(x+y)=0 and \sin(x-y)=\frac{1}{2}, where x, y \in [0, \frac{\pi}{2}], then what is the value of \cot(2x-y)?

Explanation

From the given equations, x+y = 90^\circ and x-y = 30^\circ. Solving these gives x=60^\circ and y=30^\circ. Then \cot(2x-y) = \cot(120^\circ-30^\circ) = \cot(90^\circ) = 0.

Related questions on Trigonometry

• Two poles are situated 24 m apart and their heights differ by 10 m. What is the distance between their tips?
• If \frac{\cos \theta}{1 - \sin \theta} + \frac{\cos \theta}{1 + \sin \theta} = 4, then which one of the following is a value of $(\tan^2 \...
• For 0 < \theta < \frac{\pi}{2}, consider the following : I. $(\tan^4 \theta + \tan^6 \theta)(\cot^4 \theta + \cot^6 \theta) = \sec^2 \the...
• If 3\sin \theta + 4\cos \theta = 5, then what is a value of 4\tan \theta + 3\cot \theta ?
• At a point on level ground, the tangent of the angle of elevation of the top of a tower is found to be \frac{5}{6}. On walking 70 m toward...