Question

2. Prove that: 1 + tan 30° (a) 1 -tan 30° 2+√3​

Answer 1

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

I think he thik ho?

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Answer 2

Answer:

Step-by-step explanation: