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Question

If tanx=2t1−t2 and siny=2t1+t2, then the value of dydx is

A
1
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B
t
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C
11t
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D
11+t
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Solution

The correct option is A 1
Given,
tanx=2t1t2 and siny=2t1+t2

Now,
x=tan1(2t1t2)
x=2tan1t ...... (i)
and
y=sin1(2t1+t2)
y=2tan1t ...... (ii)

From Eq. (i),
dxdt=21+t2

From Eq. (ii),
dydt=21+t2

dydx=dydt×dtdx=2(1+t2)×(1+t2)2=1

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