question_answer

Let \[0<\alpha <\pi /2\] be a fixed angle. If \[P(cos\theta ,sin\theta )\] and \[Q(cos(\alpha -\theta ),sin(\alpha -\theta )),\] then Q is obtained from P by the

A) Clockwise rotation around the origin through an angle \[\alpha \]

B) Anticlockwise rotation around the origin through an angle \[\alpha \]

C) Reflection in the line through the origin with slope \[\tan \alpha \]

D) Reflection in the line through the origin with slop \[\tan (\alpha /2)\]

Correct Answer: D

Solution :

[d] Clearly, \[OP=OQ=1,\] and \[\angle QOP=\alpha -\theta -\theta =\alpha -2\theta .\] The bisector of \[\angle QOP\] will be perpendicular to PQ and also bisect it. Hence, Q is the reflection of P in the line OM which makes an angle equal to \[\angle MOP+\angle POX\] with the x-axis, i.e., \[\frac{1}{2}(\alpha -2\theta )+\theta =\frac{\alpha }{2}\] So that slope of OM is \[\tan \,(\alpha /2)\]