1-sin60÷cos60=1-Tan30÷1+Tan30
1 answer:
Step-by-step explanation:
LHS=(1-sin60)/cos60
=(1-√3÷2)/1÷2
=2(1-√3÷2)
=2-√3
RHS=(1-tan30)/(1+tan30)
={1-(1÷√3)}/{1+(1÷√3)}
={(√3-1)/√3}/{(√3+1)/√3}
=(√3-1)/(√3+1)
={(√3-1)(√3-1)}/{(√3+1)(√3-1)}
=(3-√3-√3+1)/(3-1)
=(4-2√3)/2
=2-√3
Therefore LHS=RHS
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