pLZ help ... Q.3 and Q.4 ?
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in third
apply L hopital
-cosx/-2(π/2-x)
again it is 0/0 form again apply the rule
sinx/(-2(-1))
=1/2.....ans
(4).it is again 0/0 form... so apply the rule as above
-sinx/-1=sinx=1
apply L hopital
-cosx/-2(π/2-x)
again it is 0/0 form again apply the rule
sinx/(-2(-1))
=1/2.....ans
(4).it is again 0/0 form... so apply the rule as above
-sinx/-1=sinx=1
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