Question

Prove that tan 10+tan 70'+taniou = tanio tanfo'tam 1oo.​

Answer 1

Question:

Prove that:

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$

Answer:

Proved

Step-by-step explanation:

Given

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$

Required

Prove

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$

Subtract tan(10) from both sides

$- tan(10)+tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100) - tan(10)$

$tan(70) + tan(100) = tan(10). tan(70). tan(100) - tan(10)$

Factorize the right hand size

$tan(70) + tan(100) = -tan(10)(-tan(70). tan(100) + 1)$

Rewrite as:

$tan(70) + tan(100) = -tan(10)(1-tan(70). tan(100))$

Divide both sides by $1-tan(70). tan(100)$

$\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = \frac{-tan(10)(1-tan(70). tan(100))}{1-tan(70). tan(100))}$

$\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = -tan(10)$

In trigonometry:

$tan(A + B) = \frac{tan(A) + tan(B)}{1 - tan(A)tan(B)}$

So:

$\frac{tan(70) + tan(100)}{1 - tan(70)tan(100)}$ can be expressed as: $tan(70 + 100)$

$\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = -tan(10)$ gives

$tan(70 + 100) = -tan(10)$

$tan(170) = -tan(10)$

In trigonometry:

$tan(180 - \theta) = -tan(\theta)$

So:

$tan(180 - 10) = -tan(10)$

Because RHS = LHS

Then:

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$ has been proven