Trigonometry related questions?

Question

if A=60, B=30 verify that tan (A -- B)=

tan A - tan B

-------------------

1+tan A tan B

Answer 1

tan(A-B)= tan(60-30)= tan30= 1\sqrt3

Now,

       ( tan60-tan30)\(1+tan60 tan30)

      

       = (sqrt3-1\sqrt3)\1+ sqrt3.1\sqrt3

       =(( 3-1)\sqrt3)\ 1+1

       = (2\sqrt3)\2 = 1\sqrt3

  therefore tan(A-B)= (tanA -tanB)\1+tanA tanB

Answer 2

just put the values

at lhs you have tan30=1/sqrt3

at rhs you have tan 60-tan30/(1+tan60*tan30)

                      =(sqrt3 - 1/sqrt 3)/(1+sqrt3*1/sqrt3)=1/sqrt3

Answer 3

tan(60) - tan(30)

------------------------- =

1 + tan(60)tan(30)

sqrt(3) - 1/sqrt(3)

------------------------------- =

1 + sqrt(3) * 1/sqrt(3)

3 / sqrt(3) - 1/sqrt(3)

----------------------------- =

1 + 1

2 / sqrt(3)

-------------- =

2

1 / sqrt(3) = tan(30) = tan(60 - 30)

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Answer 4

Substitute the values, we know:

tan(A-B) = tan (60-30) = tan30 = 1/√3

tan60 = √3

So, LHS:

tan30 = 1/√3  => (1)

RHS:

(tan60 - tan30) / (1+tan60tan30)

= (√3 - 1/√3) / [1 + {√3*1/√3}]

=[(1/√3)(3-1)] / [1+1] = [2/√3]/(2) = 1/√3 => (2)

From (1) and (2)

LHS = RHS

AJM

Answer 5

A = 60 & B = 30

tan(A - B) = tan (60 -30) = tan 30 = 1/sqrt(3)

(tan A - tan B)/(1 + tan A.tan B)

= (tan 60 - tan 30)/(1 + tan 60.tan 30)

= {sqrt(3) - 1/sqrt(3)}/{1 + sqrt(3).1/sqrt(3)}

= {(3 - 1)/sqrt(3)}/(1 + 1)

= 2/sqrt(3) x 1/2

= 1/ sqrt(3)

Therefore, tan (A - B) = (tan A - tan B)/(1 + tan A.tan B)

so, This is varified.