Question

2 tanx/1 - tan2x + 1/ 2 cos2x -1= cosx + sinx/ cosx - sinx​

Answer 1

`(2 \tan x)/(1-\tan ^(2) x)+(1)/(2 \cos ^(2) x-1)=(\cos x+\sin x)/(\cos x-\sin x)` is proved

Solution:

Given that,

`(2 \tan x)/(1-\tan ^(2) x)+(1)/(2 \cos ^(2) x-1)=(\cos x+\sin x)/(\cos x-\sin x)`

Let us first solve the L.H.S

`\text { L. H.S }=(2 \tan x)/(1-\tan ^(2) x)+(1)/(2 \cos ^(2) x-1)` --- (1)

By trignometric identities,

`\tan 2 x=(2 \tan x)/(1-\tan ^(2) x)`

`\cos 2 x=2 \cos ^(2) x-1=\cos ^(2) x-\sin ^(2) x`

By using these in (1) we get,

`\text { L. H.S }=\tan 2 x+(1)/(\cos 2 x)`

By definition of tan,

`tan x = (sinx)/(cosx)`

Therefore,

`L. H.S =(\sin 2 x)/(\cos 2 x)+(1)/(\cos 2 x)\\\\L. H . S=(1+\sin 2 x)/(\cos 2 x)$` --- (ii)

By trignometric identities,

`\cos 2 x=2 \cos ^(2) x-1=\cos ^(2) x-\sin ^(2) x`

`\begin{aligned}&\sin 2 x=2 \sin x \cos x\\\\&\cos ^(2) x+\sin ^(2) x=1\end{aligned}`

By using these in (ii)

`\text { L. H.S }=(\cos ^(2) x+\sin ^(2) x+2 \sin x \cos x)/(\cos ^(2) x-\sin ^(2) x)` ----- (iii)

We know that,

`(a+b)^2 = a^2 + 2ab + b^2`

Similarly,

`(cosx + sinx)^2 = cos^2x + 2cosxsinx + sin^2x`

Also,

`a^2 - b^2 = (a+b)(a-b)`

Similarly,

`cos^2x - sin^2x = (cosx+sinx)(cosx-sinx)`

Therefore apply these in (iii)

`\begin{aligned}&\text { L. } H . S=((\cos x+\sin x)^(2))/(\cos ^(2) x-\sin ^(2) x)\\\\&\text { L. H.S }=((\cos x+\sin x)(\cos x+\sin x))/((\cos x+\sin x)(\cos x-\sin x))\end{aligned}`

Cancel out (cos x + sin x) on numerator and denominator

`\text { L. H.S }=(\cos x+\sin x)/(\cos x-\sin x)`

`(2 \tan x)/(1-\tan ^(2) x)+(1)/(2 \cos ^(2) x-1)=(\cos x+\sin x)/(\cos x-\sin x)`

Thus L.H.S = R.H.S

Thus proved