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Q.

A line OA of length r starts from its initial position OX and traces an angle AOB=α in the anticlockwise direction. It then traces back in the clockwise direction at an angle BOC - 30 (where α>3θ). Z is the foot of the perpendicular form C on OA. Also, sin3⁡θCL=cos3⁡θOL=1. 1−rcos⁡αrsin⁡α is equal to 2rsin⁡α1+2rcos⁡α is equal to2r2−1r is equal to

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a

tan⁡2θ

b

cot⁡2θ

c

sin⁡2θ

d

cos⁡2θ

e

tan2⁡θ

f

cot2⁡θ

g

cot⁡2θ

h

tan⁡2θ

i

sin⁡α

j

cos⁡α

k

sin⁡θ

l

cos⁡θ

answer is , , .

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Detailed Solution

OL=rcos⁡(α−3θ)CL=rsin⁡(α−3θ)So, sin3⁡θrsin⁡(α−3θ)=cos3⁡θrcos⁡(α−3θ)=1⇒sin3⁡θsin⁡(α−3θ)=cos3⁡θcos⁡(α−3θ)=rNow cos4⁡θ−sin4⁡θ =cos⁡θ(rcos⁡(α−3θ))−sin⁡θ(rsin⁡(α−3θ))⇒cos⁡2θ=rcos⁡(α−2θ)⇒cos⁡2θ=r(cos⁡αcos⁡2θ+sin⁡αsin⁡2θ)⇒(1−rcos⁡α)cos⁡2θ=rsin⁡αsin⁡2θ⇒1−rcos⁡αrsin⁡α=tan⁡2θ.....(i)cos3⁡θsin⁡θ+sin3⁡θcos⁡θ⇒sin⁡θcos⁡θ=rsin⁡(α−2θ)⇒sin⁡2θ=2r[sin⁡αcos⁡2θ−cos⁡αsin⁡2θ]⇒(1+2rcos⁡α)sin⁡2θ=2rsin⁡αcos⁡2θ⇒1+2rcos⁡α2rsin⁡α=cot⁡2θ.....(ii)From (1) and (2),we get1−rcos⁡αrsin⁡α=2rsin⁡α1+2rcos⁡α⇒1−rcos⁡α+2rcos⁡α−2r2cos2⁡α=2r2sin2⁡α⇒2r2−1r=cos⁡αcos4θ-sin4θ=cos⁡θ(rcos⁡(α−3θ))−sin⁡θ(rsin⁡(α−3θ))⇒cos⁡2θ=rcos⁡(α−2θ)⇒cos⁡2θ=r(cos⁡αcos⁡2θ+sin⁡αsin⁡2θ)⇒(1−rcos⁡α)cos⁡2θ=rsin⁡αsin⁡2θ⇒1−rcos⁡αrsin⁡α=tan⁡2θ.....(i)cos3⁡θsin⁡θ+sin3⁡θcos⁡θ⇒sin⁡θcos⁡θ=rsin⁡(α−2θ)⇒sin⁡2θ=2r[sin⁡αcos⁡2θ−cos⁡αsin⁡2θ]⇒(1+2rcos⁡α)sin⁡2θ=2rsin⁡αcos⁡2θ⇒1+2rcos⁡α2rsin⁡α=cot⁡2θ.....(ii)cos4θ-sin4θ=cos⁡θ(rcos⁡(α−3θ))−sin⁡θ(rsin⁡(α−3θ))⇒cos⁡2θ=rcos⁡(α−2θ)⇒cos⁡2θ=r(cos⁡αcos⁡2θ+sin⁡αsin⁡2θ)⇒(1−rcos⁡α)cos⁡2θ=rsin⁡αsin⁡2θ⇒1−rcos⁡αrsin⁡α=tan⁡2θ.....(i)cos3⁡θsin⁡θ+sin3⁡θcos⁡θ⇒sin⁡θcos⁡θ=rsin⁡(α−2θ)⇒sin⁡2θ=2r[sin⁡αcos⁡2θ−cos⁡αsin⁡2θ]⇒(1+2rcos⁡α)sin⁡2θ=2rsin⁡αcos⁡2θ⇒1+2rcos⁡α2rsin⁡α=cot⁡2θ.....(ii)from (i) and (ii)1−rcos⁡αrsin⁡α=2rsin⁡α1+2rcos⁡α⇒1−rcos⁡α+2rcos⁡α−2r2cos2⁡α=2r2sin2⁡α⇒2r2−1r=cos⁡α
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