Question

Can anyone provide me the right answer :-

${\frac{tan^2 2x - tan^2x }{1-tan^2 2x\; tan^2x }= tan3x \; tanx}$​

Answer 1

Here's we are asked to prove :-

• $\sf\purple{{\dfrac{tan^2 2x - tan^2x }{1-tan^2 2x\; tan^2x }= tan3x \; tanx}}$

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Formula's:-

✝$\sf\:a^2-b^2=(a+b)(a-b)$

✝$\sf{tan(a+b)=\dfrac{tan \;a+tan\;b}{1-tan \;a\;tan \;b}\;\;}$

✝$\sf{tan(a-b)=\dfrac{tan\;a-tan\;b}{1+tan\;a\;tan\;b}}$

Taking LHS

$\sf\red{ :\implies \dfrac{tan^2 2x - tan^2x}{1-tan^22x \; tan^2x}}$

$\sf :\implies\dfrac{(tan2x+tanx) (tan2x-tanx)}{(1-tan2x\;tanx)(1+tan2x\;tanx)}$

$\sf :\implies \left(\dfrac{tan2x+tanx}{1-tan2x\;tanx}\right) \left( \dfrac{tan2x-tanx}{1+tan2x\;tanx}\right)$

$\sf :\implies tan(2x+x) \; tan(2x-x)$

$\sf \red {:\implies tan3x \;tanx \; }= RHS$

$\:\:\:$

• (Proved..!!)

Answer 2

Step-by-step explanation:

$see \: above \: u \: will \: get \: the \: right \: answer$

$\blue { {\frac{tan^2 2x - tan^2x }{1-tan^2 2x\; tan^2x }= tan3x \; tanx}}$