Question

1-sin60÷cos60=1-Tan30÷1+Tan30

Answer 1

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