arctan(z)=12ilog(1+iz1−iz)
i have tried but my answer doesn't matches to the equation .the componendo dividendo property might have been used. where
arcsin(x)=1ilog(iz+√1−z2)
How to find limh→0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
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