Question

Prove Pls I will mark the best answer as brainliest

Answer 1

To prove :

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

Proof :

Taking LHS

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

$\star\pink{\sf{(using\:identity\:a^2-b^2=(a+b)(a-b))}}$

$\red{\implies}\sf{LHS=\dfrac{(tan2x+tanx) (tan2x-tanx)}{(1-tan2x\;tanx)(1+tan2x\;tanx)}}$

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

$\star\pink{\sf{using\;trigonometric\;identities}}$

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

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

$\red{\implies}\sf{LHS=tan(2x+x) \; tan(2x-x)}$

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

$\:\:\:$

Proved.