if theta = 60° , show that tan theta = root 1-cos square theta / cos theta
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α=60
tanα=√3=RHS
Now, √1-cos^α =sinα =√3/2
cosα=1/2
(√1-cos. ^2α)/cosα=(√3/2)/(1/2)=√3=LHS
LHS=RHS...HENCE PROVED
tanα=√3=RHS
Now, √1-cos^α =sinα =√3/2
cosα=1/2
(√1-cos. ^2α)/cosα=(√3/2)/(1/2)=√3=LHS
LHS=RHS...HENCE PROVED
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