Question

Guys please answer this quickly I need help!

Answer 1

Step-by-step explanation:

Assume two angles A and B

let

A + B =45

taking tan on both sides

tan(A+B) = tan45

Applying the formula

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

tanA + tan B = 1-tanA.tanB

tan A + tanB + tanA.tanB= 1 --------(1)

According to question::-

(1+tan6 1/2)(1+tan 38 1/2)

= (1+ tan 13/2)(1+ tan 77/2)

= tan 13/2 + tan 77/2 + tan 13/2.tan 77/2 +1 -----------(2)

Applying the formula we got in step. (1) as {(13+77)/2=45}

putting in step (2) we get

= 1+1

= 2

Hence the answer is 2.

Answer 2

We need to know the equation

$tan(x+y) = \frac{tanx+tany}{1-tanxtany}\\\\ \large\black\boxed{tanx+tany = tan(x+y)(1-tanxtany)}$

$(1+tan6\frac{1}{2})(1+tan38\frac{1}{2})\\ \\= 1 + tan6\frac{1}{2}+tan38\frac{1}{2} + ( tan6\frac{1}{2} \times tan38\frac{1}{2})\\ \\= (1 + tan6\frac{1}{2} \times tan38\frac{1}{2})+ (tan6\frac{1}{2}+tan38\frac{1}{2})\\\\=1+tan6\frac{1}{2}tan38\frac{1}{2} + tan(6\frac{1}{2}+38\frac{1}{2}) \times (1- tan6\frac{1}{2}tan38\frac{1}{2})\\\\= 1+tan6\frac{1}{2}tan38\frac{1}{2} + tan(45) \times (1- tan6\frac{1}{2}tan38\frac{1}{2})$

$= 1+tan6\frac{1}{2}tan38\frac{1}{2}+1-tan6\frac{1}{2}tan38\frac{1}{2}\\ \\=2$