Question:
Prove that:
[tex]tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)[/tex]
Answer:
Proved
Step-by-step explanation:
Given
[tex]tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)[/tex]
Required
Prove
[tex]tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)[/tex]
Subtract tan(10) from both sides
[tex]- tan(10)+tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100) - tan(10)[/tex]
[tex]tan(70) + tan(100) = tan(10). tan(70). tan(100) - tan(10)[/tex]
Factorize the right hand size
[tex]tan(70) + tan(100) = -tan(10)(-tan(70). tan(100) + 1)[/tex]
Rewrite as:
[tex]tan(70) + tan(100) = -tan(10)(1-tan(70). tan(100))[/tex]
Divide both sides by [tex]1-tan(70). tan(100)[/tex]
[tex]\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = \frac{-tan(10)(1-tan(70). tan(100))}{1-tan(70). tan(100))}[/tex]
[tex]\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = -tan(10)[/tex]
In trigonometry:
[tex]tan(A + B) = \frac{tan(A) + tan(B)}{1 - tan(A)tan(B)}[/tex]
So:
[tex]\frac{tan(70) + tan(100)}{1 - tan(70)tan(100)}[/tex] can be expressed as: [tex]tan(70 + 100)[/tex]
[tex]\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = -tan(10)[/tex] gives
[tex]tan(70 + 100) = -tan(10)[/tex]
[tex]tan(170) = -tan(10)[/tex]
In trigonometry:
[tex]tan(180 - \theta) = -tan(\theta)[/tex]
So:
[tex]tan(180 - 10) = -tan(10)[/tex]
Because RHS = LHS
Then:
[tex]tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)[/tex] has been proven
