How do I prove that $\frac{2}{cot x tan 2 x} \equiv 1 - {tan}^{2} x$ $2/(cotxtan2x) -= 1 - tan^2x$ ?

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Answer 1 · 2018-06-20T14:09:05

As proved below.

$cot x = \frac{1}{tan x}$ $cot x = 1/ tan x$

$tan 2 x = \frac{2 tan x}{1 - {tan}^{2} x}$ $tan 2x = (2 tan x) / (1 - tan^2 x)$ - Identity.

To prove $\frac{2}{cot x \cdot tan 2 x} = 1 - {tan}^{2} x$ $2 / (cot x * tan 2x) = 1 - tan^2 x$

$\frac{2}{cot x \cdot tan 2 x}$ $2 / (cot x * tan 2x)$

=> (2 tan x) / tan 2x#

$\Rightarrow \frac{2 tan x}{\frac{2 tan x}{1 - {tan}^{2} x}}$ $=> (2 tan x) / ((2tan x) / (1 - tan^2 x))$

$\Rightarrow (1 - {tan}^{2} x) = R H S$ $=> (1 - tan^2 x) = R H S$

How do I prove that 2cotxtan2x≡1−tan2x2/(cotxtan2x) -= 1 - tan^2x?