Question

(1 + sin 2x - cos 2 x)/
(1 + sin 2 x + cos 2 x) =
tan x Prove

Answer 1

Answer:

LHS=RHS

Step-by-step explanation:

sin2x=2sinxcosx

cos2x=cos^2x-sin^2x

1-cos2x=(cos^2x+sin^2x)-(cos^2x-sin^2x)

=2sin^2x

1+cos2x=(cos^2x+sin^2x)+(cos^2x-sin^2x)

=2cos^2x

LHS=(2sin^2x+2sinxcosx)/(2cos^2x+2sinxcosx)

=(2sinx(sinx+cosx))/(2cosx(cosx+sinx))

=2sinx/2cosx

=sinx/cosx

=tanx =RHS

Hence proved.

Answer 2

To Prove :-

$\sf \dfrac{1+sin2x-cos2x}{1+sin2x+cos2x}= tanx$

Solution :-

$\sf L.H.S = \dfrac{1+sin2x-cos2x}{1+sin2x+cos2x}$

 $\bullet \bf \ sin2x = 2sinx cosx \\\\\bullet \ cos2x = cos^2-sin^2x$

         $= \sf \dfrac{1+2sinxcosx-cos^2x+sin^2x}{1+2sinxcosx+cos^2x-sin^2x} \\\\= \dfrac{(1-cos^2x)+2sinxcosx+sin^2x}{(1-sin^2x)+2sinxcosx+cos^2x}$

$\bullet \bf \ 1-cos^2x = sin^2x \\\\\bullet \ 1-sin^2x = cos^2x$

         $= \sf \dfrac{sin^2x+2sinxcosx+sin^2x}{cos^2x+2sinxcosx+cos^2x}\\\\= \dfrac{2sin^2x+2sinxcosx}{2cos^2x+2sinxcosx}\\\\= \dfrac{2sinx(sinx+cosx)}{2cosx(cosx+sinx)}\\\\= \dfrac{sinx}{cosx} \\\\= tanx\\\\= R.H.S$

Hence proved.

Other identities :-

$\bullet \bf \ cos2x = 2cos^2-1 \\\\\bullet \ cos2x = 1-2sin^2x\\\\\bullet \ cos2x = \dfrac{1-tan^2x}{1+tan^2x}$

$\bullet \bf \ sin2x = \dfrac{2tanx}{1+tan^2x}$