# what is the answer of proving the identity of cos2 x tan x=2sin x/sec x+sin2 x(sec x) please help pls....tnx

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`cos 2x tan x = (2sin x)/(sec x) + sin 2x(sec x)`

Note that, `sec x = 1/ (cos x)` .

`cos 2x tan x = 2sinx cos x + (sin 2x)/cos x`

To simplify right side, use the double angle identity of sine function which is, sin 2A = 2sin A cos A .

`cos 2x tan x = 2sin x cos x + (2sin x cos x)/(cos x)`

`cos2x tanx = 2sin x cos x + 2sin x`

`cos 2x tan x = 2sinx ( cos x+ 1)`

Note that, `tanx = (sinx)/(cosx)` .

`cos 2x (sin x)/(cosx) = 2sin x (cos x + 1)`

Multiply both sides by `(cosx)/(sinx)` to simplify the expression.

`cos 2x = 2 cos x(cos x + 1)`

`cos 2x = 2cos^2x + 2cos x`

Then, expand left side using the double angle identity of cosine which is `cos2A = 2cos^2A - 1` .

`2cos^2x - 1 = 2cos^2 x + 2cos x`

`-1 = 2cos^2x - 2cos^2x + 2cos x`

`-1 = 2cos x`

`-1/2 = cos x` ` `

From here, refer to Unit Circle Chart or Table of Trigonometric Functions for Special Angles to determine the value of angle x.

Base on it, x = 120 deg., and 240 deg.

Since there is no indicated interval for x, the general solution is:

`x = 120 + 360k ` deg. and `x = 240 + 360k` deg.

**Answer: The above equation will only be true if the values of x are 120+360k degrees and 240 + 360k degrees**.