Question #db1ea

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Question #db1ea

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Answer 1 · 2017-03-15T04:15:03

see explanation below

Explanation:

Let we take LHS to prove RHS.

$tan (x + \frac{3.14}{4}) = tan (x + \frac{π}{4})$ $tan (x + 3.14/4) = tan(x + pi/4)$

from Phytogoras Triginometric method,

$tan (a + b) = \frac{tan a + tan b}{1 - tan a \cdot tan b}$ $tan ( a + b) = (tan a + tan b)/(1 - tan a* tan b)$ ,

therefore
$tan (x + \frac{π}{4}) = \frac{tan x + tan (\frac{π}{4})}{1 - tan x \cdot tan (\frac{π}{4})}$ $tan(x + pi/4) = (tan x + tan (pi/4))/(1 - tan x* tan (pi/4)$

$tan (x + \frac{π}{4}) = \frac{tan x + 1}{1 - tan x \cdot 1}$ $tan(x + pi/4) = (tan x + 1)/(1 - tan x* 1)$ where $tan (\frac{π}{4}) = 1$ $tan (pi/4) = 1$

$tan (x + \frac{π}{4}) = \frac{tan x + 1}{1 - tan x}$ $tan(x + pi/4) = (tan x + 1)/(1 - tan x)$ or $\frac{1 + tan x}{1 - tan x}$ $(1 + tan x)/(1 - tan x)$

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Question #db1ea

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