Respuesta :
[tex]\frac{2 \tan x}{1-\tan ^{2} x}+\frac{1}{2 \cos ^{2} x-1}=\frac{\cos x+\sin x}{\cos x-\sin x}[/tex] is proved
Solution:
Given that,
[tex]\frac{2 \tan x}{1-\tan ^{2} x}+\frac{1}{2 \cos ^{2} x-1}=\frac{\cos x+\sin x}{\cos x-\sin x}[/tex]
Let us first solve the L.H.S
[tex]\text { L. H.S }=\frac{2 \tan x}{1-\tan ^{2} x}+\frac{1}{2 \cos ^{2} x-1}[/tex] --- (1)
By trignometric identities,
[tex]\tan 2 x=\frac{2 \tan x}{1-\tan ^{2} x}[/tex]
[tex]\cos 2 x=2 \cos ^{2} x-1=\cos ^{2} x-\sin ^{2} x[/tex]
By using these in (1) we get,
[tex]\text { L. H.S }=\tan 2 x+\frac{1}{\cos 2 x}[/tex]
By definition of tan,
[tex]tan x = \frac{sinx}{cosx}[/tex]
Therefore,
[tex]L. H.S =\frac{\sin 2 x}{\cos 2 x}+\frac{1}{\cos 2 x}\\\\L. H . S=\frac{1+\sin 2 x}{\cos 2 x}$[/tex] --- (ii)
By trignometric identities,
[tex]\cos 2 x=2 \cos ^{2} x-1=\cos ^{2} x-\sin ^{2} x[/tex]
[tex]\begin{aligned}&\sin 2 x=2 \sin x \cos x\\\\&\cos ^{2} x+\sin ^{2} x=1\end{aligned}[/tex]
By using these in (ii)
[tex]\text { L. H.S }=\frac{\cos ^{2} x+\sin ^{2} x+2 \sin x \cos x}{\cos ^{2} x-\sin ^{2} x}[/tex] ----- (iii)
We know that,
[tex](a+b)^2 = a^2 + 2ab + b^2[/tex]
Similarly,
[tex](cosx + sinx)^2 = cos^2x + 2cosxsinx + sin^2x[/tex]
Also,
[tex]a^2 - b^2 = (a+b)(a-b)[/tex]
Similarly,
[tex]cos^2x - sin^2x = (cosx+sinx)(cosx-sinx)[/tex]
Therefore apply these in (iii)
[tex]\begin{aligned}&\text { L. } H . S=\frac{(\cos x+\sin x)^{2}}{\cos ^{2} x-\sin ^{2} x}\\\\&\text { L. H.S }=\frac{(\cos x+\sin x)(\cos x+\sin x)}{(\cos x+\sin x)(\cos x-\sin x)}\end{aligned}[/tex]
Cancel out (cos x + sin x) on numerator and denominator
[tex]\text { L. H.S }=\frac{\cos x+\sin x}{\cos x-\sin x}[/tex]
[tex]\frac{2 \tan x}{1-\tan ^{2} x}+\frac{1}{2 \cos ^{2} x-1}=\frac{\cos x+\sin x}{\cos x-\sin x}[/tex]
Thus L.H.S = R.H.S
Thus proved
