AP is a diameter of a unit circle with centre at O. Let AC be an arc of this circle…
Mathematics · JEE Advanced · NTA Exams — Limit, Continuity and Differentiability
AP is a diameter of a unit circle with centre at O. Let AC be an arc of this circle, which subtends angle θ radian at centre O. A tangent line is drawn to the circle at the point A and a segement AB on this tangent is laid off whose length is equal to that of the arc AC. A straight line BC is drawn to intersect the extension of the diameter AP at Q. CD is the perpendicular that fall from the point C upon the diameter AP.
The length AQ equal to
- \(\frac{\theta(1-\cos \theta)}{\theta-\sin \theta}\)
- \(\frac{\theta(1-\cos \theta)}{\theta+\sin \theta}\)
- \(\frac{\theta(1+\cos \theta)}{\theta-\sin \theta}\)
- \(\frac{\theta(1+\cos \theta)}{\theta+\sin \theta}\)
Answer
(A) (1- ) -
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