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Question

# If ex=√1+t−√1−t√1+t+√1−t and tany2=√1−t1+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 −12Let t=cos2θ ex=√1+cos2θ−√1−cos2θ√1+cos2θ+√1−cos2θ⇒ex=cosθ−sinθcosθ+sinθ⇒ex=1−tanθ1+tanθ=tan(π4−θ)⇒exdxdθ=−sec2(π4−θ) tany2=√1−cos2θ1+cos2θ⇒tany2=tanθ⇒12sec2y2dydθ=sec2θ When t=12=cos2θ⇒θ=π6 ex=tan(π4−π6)⇒x=lntanπ12 tany2=tanπ6⇒y=π3 So, dydx=dydθdxdθ⇒dydx=2cos2y22sec2θ−e−xsec2(π4−θ) Putting t=12,x=lntanπ12,y=π3 ⇒dydx=2cos2π62sec2π6−e−lntanπ/12sec2(π12)⇒dydx=2−cotπ12sec2(π12)⇒dydx=−2tanπ12cos2π12⇒dydx=−sinπ6=−12

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