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If ex=1+t-1-t1+t+1-t and tan y/2=1-t1+t then dydx at t=1/2 is

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detailed solution

Correct option is A

ex=1-tan y21+tan y2=tan π4-y2 y=2π4-tan-1 ex dydx=2-11+ex2ex  At t=12,tany2=1-121+12=13y2=π6 y=π3, ex=3-13+1=2-3 dydx=-2(2-3)1+(2-3)2=-2(2-3)8-43=-24=-12 

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