Question

(1+tan 18°)(1 + tan 27) =​

Answer 1

Answer:

Tan(A+B)=tanA+tanB/1-tanAtanB

tanA+tanB= tan(A+B)*(1-tanAtanB)

tan18+tan27=tan(18+27)*(1-tan18*tan27)

tan18+tan27=tan45*(1-tan18*tan27)

tan18+tan27 =1-tan18*tan27

tan18+tan27+tan18*tan27=1

Answer 2

The value of (1 + tan 18°)(1 + tan 27°) is 2.

Step-by-step explanation:

Given: (1 + tan 18°)(1 + tan 27°)

To Find: Value of the given expression

Solution:

• Finding the value of (1 + tan 18°)(1 + tan 27°) is 2

⇒ (1 + tan 18°)(1 + tan 27°) = 1 + tan 18° + tan 27° + tan 18°tan 27° . . . . . . (1)

Now, considering the term (tan 18° + tan 27°) such that using the identity, $tan (A+B) = \dfrac{tanA + tanB}{1-tanAtanB} \\\\\Rightarrow tanA+tanB = tan(A+B) \times (1-tanAtanB)$

we get,

⇒ tan 18° + tan 27° = tan (18° + 27°) × (1-tan 18°tan 27°)

⇒ tan 18° + tan 27° = tan (45°) × (1-tan 18°tan 27°)

⇒ tan 18° + tan 27° = 1 × (1-tan 18°tan 27°)

⇒ tan 18° + tan 27° = (1-tan 18°tan 27°) . . . . . (2)

Substituting (2) in (1), we get,

⇒ (1 + tan 18°)(1 + tan 27°) = 1 + 1 - tan 18°tan 27° + tan 18°tan 27°

⇒ (1 + tan 18°)(1 + tan 27°) = 1 + 1 = 2

Hence, the value of (1 + tan 18°)(1 + tan 27°) is 2.