Answer:
[tex]sinx+cosx=\frac{cosx}{1-tanx}+\frac{sinx}{1-cotx}\\[/tex] proved.
Step-by-step explanation:
[tex]sinx+cosx=\frac{cosx}{1-tanx}+\frac{sinx}{1-cotx}\\[/tex]
Taking R.H.S
[tex]\frac{cosx}{1-tanx}+\frac{sinx}{1-cotx}\\[/tex]
Multiply and divide first term by cos x and second term by sinx
[tex]=\frac{cosx*cosx}{cosx(1-tanx)}+\frac{sinx*sinx}{sinx(1-cotx)}[/tex]
we know tanx = sinx/cosx and cotx = cosx/sinx
[tex]=\frac{cos^2x}{cosx(1-\frac{sinx}{cosx} )}+\frac{sin^2x}{sinx(1-\frac{cosx}{sinx})}\\=\frac{cos^2x}{cosx-sinx}+\frac{sin^2x}{sinx-cosx}[/tex]
Taking minus(-) sign common from second term
[tex]=\frac{cos^2x}{cosx-sinx}-\frac{sin^2x}{cosx-sinx}[/tex]
taking LCM of cosx-sinx and cosx-sinx is cosx-sinx
[tex]=\frac{cos^2x-sin^2x}{cosx-sinx}[/tex]
We know a^2-b^2 = (a+b)(a-b), Applying this formula:
[tex]=\frac{(cosx+sinx)(cosx-sinx)}{cosx-sinx}\\=cosx+sinx\\=L.H.S[/tex]
Hence proved
