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Question

If ex=1+t1t1+t+1t and tany2=1t1+t then dydx at t=12 is

A
12
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B
12
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C
0
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D
None of these
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Solution

The correct option is B 12
Let t=cos2θ
ex=1+cos2θ1cos2θ1+cos2θ+1cos2θex=cosθsinθcosθ+sinθex=1tanθ1+tanθ=tan(π4θ)exdxdθ=sec2(π4θ)

tany2=1cos2θ1+cos2θtany2=tanθ12sec2y2dydθ=sec2θ

When t=12=cos2θθ=π6
ex=tan(π4π6)x=lntanπ12
tany2=tanπ6y=π3

So,
dydx=dydθdxdθdydx=2cos2y22sec2θexsec2(π4θ)
Putting t=12,x=lntanπ12,y=π3
dydx=2cos2π62sec2π6elntanπ/12sec2(π12)dydx=2cotπ12sec2(π12)dydx=2tanπ12cos2π12dydx=sinπ6=12

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