Question

lf tan(A-B)=tanA-tanB/1+tanA tanB then find tan 15°​

Answer 1

Answer:

2-√3

Step-by-step Explanation:

Given:

tan(A-B) = tanA-tanB/1+tanA tanB

Solution:

tan15°

tan15°= tan(45°-30°)

tan15°= tan45°-tan30°/1+tan45° tan30°

[As tan(A-B) = tanA-tanB/1+tanA tanB]

= (1-1/√3)/1+1×1/√3

= (√3-1)/(√3+1)

= 2-√3

tan15° = 2-√3

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Answer 2

Step-by-step explanation:

Tan(A-B) = TanA - TanB / 1 + TanA TanB

Tan15° = Tan(60°-45°) = Tan60° - Tan45°/ 1 + Tan60°.Tan45° -------- (1)

Value of Tan60° = √3

Value of Tan45° = 1

Substituting the values of Tan60° and Tan45° in eq.1

= √3- 1 / 1 + √3*1

Tan15°= √3 - 1/1 +√3

Rationalising denominator, we get

Tan15° = √3 - 1/ 1 +√3 x 1 - √3 / 1 -√3

Tan15° = √3 + √3 - 1 - √9 / 1 - √9

= 2√3 - 4 / -2

Tan15° = -(√3 - 2 )

Tan15° = 2 - √3