If $3 tan 2 x - 4 tan 3 x = {tan}^{2} 3 x tan 2 x$ then $tan 3 x + 3 tan x + 7$ is equal to

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Solution

$3 tan 2 x - 4 tan 3 x = {tan}^{2} 3 x tan 2 x$

$3 tan 2 x - 3 tan 3 x = tan 3 x + {tan}^{2} 3 x tan 2 x$

$\Rightarrow 3 (tan 2 x - tan 3 x) = tan 3 x (1 + tan 3 x tan 2 x)$

$\Rightarrow 3 \frac{tan 2 x - tan 3 x}{1 + tan 3 x tan 2 x} = tan 3 x$

$\Rightarrow - 3 tan x = tan 3 x$

Hence $tan 3 x + 3 tan x + 7 = 7$

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Similar questions

3 tan 2 x - 4 tan 3 x = {tan}^{2} 3 x tan 2 x

3 t a n 2 x - 4 t a n 3 x = t a n^{2} 3 x t a n 2 x

t a n 3 x = \frac{3 t a n x - t a n^{3} x}{1 - 3 t a n^{2} x}

t a n x + t a n (x + \frac{π}{3}) + t a n (x + \frac{2 π}{3}) = 3,

\frac{3 t a n x - t a n^{3} x}{1 - 3 t a n^{2} x} = 1

4 {tan}^{3} x - 3 tan x = 0

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