Question

2 x
4tan x (1 - tanºx)
tan 4x =
1-6 tan” x + tan* x
4​

Answer 1

Answer:

$\bf \large\underbrace\orange{Question \: Correct:}$

$\bf \:To \: prove :- tan4x =\dfrac{2tanx(1 - {tan}^{2} x)}{1 + {tan}^{4} x - 6 {tan}^{2} x}$

$\bf \large\underbrace\orange{Answer:}$

$\bf\underbrace\orange{Identity Used :}$

$\bf \:tan2x = \dfrac{2tanx}{1 - {tan}^{2} x}$

$\bf\underbrace\orange{Solution:}$

Consider LHS

$\bf \:tan4x$

$\bf\implies \:tan2(2x)$

$\bf\implies \:\dfrac{2tan2x}{1 - {tan}^{2} 2x}$

$\bf\implies \:\dfrac{\dfrac{2tanx}{1 - {tan}^{2} x} }{1 - {(\dfrac{2tanx}{1 - {tan}^{2} x}) }^{2} }$

$\bf\implies \:\dfrac{2tan}{1 - {tan}^{2} x} \times \dfrac{ {(1 - {tan}^{2}x) }^{2} }{ {(1 - {tan}^{2} x)}^{2} - 4 {tan}^{2}x }$

$\bf\implies \:\dfrac{2tanx(1 - {tan}^{2} x)}{1 + {tan}^{4}x - 2 {tan}^{2} x - 4 {tan}^{2}x }$

$\bf\implies \:\dfrac{2tanx(1 - {tan}^{2} x)}{1 + {tan}^{4} x - 6 {tan}^{2} x}$

_____________________________________________