How do you prove $[(2tanx)/(1-tan^2x)]+[(1)/(2cos^2x-1)]= (cosx+sinx)/(cosx-sinx)$ ?

Question

15054 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-26T13:55:33

Please see below.

We know that,

$color(red)((1)tan2x=(2tanx)/(1-tan^2x)$

$color(red)((2)cos2x=2cos^2x-1=cos^2x-sin^2x$

$color(red)((3)sin2x=2sinxcosx,and cos^2x+sin^2x=1$

We have,

$(2tanx)/(1-tan^2x)+1/(2cos^2x-1)=(cosx+sinx)/(cosx-sinx)$

Now,

$LHS=(2tanx)/(1-tan^2x)+1/(2cos^2x-1$

$=tan2x+1/(cos2x).......to$ Using $(1) and (2)$

$=(sin2x)/(cos2x)+1/(cos2x)$

$=(1+sin2x)/(cos2x)$

$=(sin^2x+cos^2x+2sinxcosx)/(cos^2x-sin^2x).....to$ Using $(2) and (3)$

$=(cosx+sinx)^cancel(2)/(cancel((cosx+sinx))(cosx-sinx))$

$=(cosx+sinx)/(cosx-sinx)$

$=RHS$

How do you prove [(2tanx)/(1-tan^2x)]+[(1)/(2cos^2x-1)]= (cosx+sinx)/(cosx-sinx)[(2tanx)/(1-tan^2x)]+[(1)/(2cos^2x-1)]= (cosx+sinx)/(cosx-sinx)?