What is tan20tan32 plus tan32tan38 plus tan38tan20?

Answer 1

Oh honey, you're throwing some trigonometry at me? Alright, buckle up. The sum of tan20tan32 plus tan32tan38 plus tan38tan20 is equal to 1. Just plug in those values and watch the magic happen. Math can be sassy too, you know.

Answer 2

The expression tan20tan32 + tan32tan38 + tan38tan20 can be simplified using the tangent addition formula: tan(A + B) = (tanA + tanB) / (1 - tanAtanB). By applying this formula, we can simplify the expression to tan(20+32) + tan(32+38) + tan(38+20), which equals tan52 + tan70 + tan58. Further simplification using trigonometric identities may be needed to obtain a final numerical result.

Answer 3

Oh, what a happy little math problem we have here! To solve this, we can use the formula for the sum of tangents of two angles. We can simplify the expression to tan(20 + 32) + tan(32 + 38) + tan(38 + 20), which equals tan(52) + tan(70) + tan(58). Finally, we can calculate the values of tan(52), tan(70), and tan(58) to find the sum.

Answer 4

This may not be the most efficient method but ...

Let the three angle be A, B and C.

Then note that A + B + C = 20+32+38 = 90

so that C = 90-A+B.

Therefore,

sin(C) = sin[(90-(A+B) = cos(A+B)

and cos(C) = cos[(90-(A+B) = sin(A+B).

So that tan(C) = sin(C)/cos(C) = cos(A+B) / sin(A+B) = cot(A+B)

Now, tan(A+B) = [tan(A)+tan(B)] / [1- tan(A)*tan(B)]

so cot(A+B) = [1- tan(A)*tan(B)] / [tan(A)+tan(B)]

The given expressin is

tan(A)*tan(B) + tan(B)*tan(C) + tan(C)*tan(A)

= tan(A)*tan(B) + [tan(B) + tan(A)]*cot(A+B)

substituting for cot(A+B) gives

= tan(A)*tan(B) + [tan(B) + tan(A)]*[1- tan(A)*tan(B)]/[tan(A)+tan(B)]

cancelling [tan(B) + tan(A)] and [tan(A) + tan(B)], which are equal, in the second expression.

= tan(A)*tan(B) + [1- tan(A)*tan(B)]

= 1