prove: tan10+tan70+tan100=tan10.tan70.tan100
100+70 = -10
Adding 'tan' on both sides,
tan(100+70)= -tan10
(tan100+tan70)/1-tan100.tan70 = -tan10
Multiplying 1-tan100.tan70 with -tan10,
tan100+ tan70= -tan10+tan10.tan100.tan70
tan10+tan100+tan70= tan10.tan70.tan100 proved
How 100-70=-10
Haven't tried it, but look at this:
https://answers.yahoo.com/question/index?qid=20150724191938AAydOpZ
sure looks messy, I am sure there is an easier way
100+70=170
So 100+10+70=180
170-180=-10
To prove the equation, tan10 + tan70 + tan100 = tan10 * tan70 * tan100, we will use the properties of the tangent function, known trigonometric identities, and algebraic manipulations.
Step 1: Express tan(10), tan(70), and tan(100) using their respective angle sum and difference identities.
We can use the identity tan(A + B) = (tanA + tanB) / (1 - tanA * tanB).
Using this identity:
tan(10) = tan(60 - 50) = (tan60 + tan(-50)) / (1 - tan60 * tan(-50))
Note that tan(-50) = -tan(50), as tangent is an odd function.
tan(10) = (√3 + (-tan50)) / (1 - √3 * (-tan50))
= (√3 - tan50) / (1 + √3 * tan50)
Similarly, using the same identity:
tan(70) = tan(60 + 10) = (tan60 + tan10) / (1 - tan60 * tan10)
= (√3 + tan10) / (1 - √3 * tan10)
tan(100) = tan(150 - 50) = (tan150 + tan(-50)) / (1 - tan150 * tan(-50))
= (-√3 + (-tan50)) / (1 + √3 * tan50)
= (tan50 - √3) / (1 + √3 * tan50)
Step 2: Substitute the derived values into the original equation.
tan10 + tan70 + tan100 = (√3 - tan50) / (1 + √3 * tan50) + (√3 + tan10) / (1 - √3 * tan10) + (tan50 - √3) / (1 + √3 * tan50)
Step 3: Simplify the expression by obtaining a common denominator and combining the fractions.
Since the denominators are already the same, we can combine the numerators:
= [(√3 - tan50)(1 - √3 * tan10) + (√3 + tan10)(1 + √3 * tan50) + (tan50 - √3)(1 - √3 * tan10)] / (1 + √3 * tan50)
= [(√3 - tan50 - √3 * tan50 + √3 * tan10^2) + (√3 + √3 * tan10 + tan10 * √3 * tan50 + tan10 * √3)] / (1 + √3 * tan50)
= [(√3 - √3 * tan50 - √3 * tan50 + √3 * tan100) + (√3 + √3 * tan10 + tan10 * √3 * tan50 + tan10 * √3)] / (1 + √3 * tan50)
= (2√3) / (1 + √3 * tan50)
Step 4: Simplify further using the identity tan(A) = sin(A) / cos(A).
= (2√3) / (1 + √3 * (sin50 / cos50))
= (2√3 * cos50) / (cos50 + √3 * sin50)
Step 5: Use the identity sin(a + b) = sin(a) * cos(b) + cos(a) * sin(b), and cos(a + b) = cos(a) * cos(b) - sin(a) * sin(b).
= (2√3 * cos50) / (√(3/4) * (cos50 + √3/2 * sin50))
= (2√3 * cos50) / (√3/2 * (cos50 + √3/2 * sin50))
= 2 * 2 * cos50 / (2 * √3/2)
= 2 * cos50 / √3
= 2 * (cos50 * √3) / 3
= 2 * sin(90 - 50) / 3
= 2 * sin40 / 3
Now, the further simplification is not possible because sin40 cannot be expressed in terms of simple radicals or rational numbers.
Therefore, we conclude that tan10 + tan70 + tan100 is not equal to tan10 * tan70 * tan100.