Prove:
(tanx)(sinx) / (tanx) + (sinx) = (tanx) - (sinx) / (tanx)(sinx)
What I have so far:
L.S.
= (sinx / cosx) sinx / (sinx / cosx) + sinx
= (sin^2x / cosx) / (sinx + (sinx) (cosx) / cosx)
= (sin^2x / cosx) / (cosx / sinx + sinxcosx)
your equation is not an identity the way you wrote it,
try substituting any angle into the equation, it will not satisfy the equation.
The way you wrote it,
LS reduces to simply 2sinx
use brackets to identify the exact order of operation.
To prove the equation (tanx)(sinx) / (tanx) + (sinx) = (tanx) - (sinx) / (tanx)(sinx), we need to simplify both the left-hand side (LHS) and the right-hand side (RHS) and show that they are equal.
Starting with the LHS:
(tanx)(sinx) / (tanx) + (sinx)
1. Rewrite tanx in terms of sinx and cosx:
(tanx)(sinx) = (sinx / cosx)(sinx) = (sin^2x / cosx)
2. Rewrite (tanx) + (sinx) as a common denominator:
(tanx) + (sinx) = (sinx/cosx) + (sinx) = [(sinx + sinx*cosx) / cosx]
3. Substitute the rewritten forms into the equation:
LHS = (sin^2x / cosx) / [(sinx + sinx*cosx) / cosx]
To simplify further, we can multiply the numerator and denominator by cosx to remove the fraction within a fraction:
LHS = (sin^2x / cosx) * (cosx / (sinx + sinx*cosx))
= sin^2x / (sinx + sinx*cosx)
Moving on to the RHS:
(tanx) - (sinx) / (tanx)(sinx)
1. Rewrite tanx in terms of sinx and cosx:
(tanx) = (sinx / cosx)
2. Rewrite (tanx)(sinx) as a common denominator:
(tanx)(sinx) = (sinx / cosx)(sinx) = (sin^2x / cosx)
3. Rewrite (tanx)(sinx) - (sinx) as a common denominator:
(tanx)(sinx) - (sinx) = (sinx / cosx)(sinx) - (sinx) = [(sin^2x - sinx*cosx) / cosx]
4. Rewrite the denominator as (tanx)(sinx):
(tanx)(sinx) = (sinx / cosx)(sinx) = (sin^2x / cosx)
Substituting the rewritten forms into the equation:
RHS = (sin^2x - sinx*cosx) / [(sin^2x / cosx)(sinx)]
To simplify further, we can multiply the numerator and denominator by cosx*sinx to remove the fraction within a fraction:
RHS = (sin^2x - sinx*cosx) * (cosx*sinx) / (sin^2x)
= (sin^2x - sinx*cosx) * (cosx*sinx) / (sinx*sinx)
= (sin^2x - sinx*cosx) * cosx
Now, we need to show that the LHS is equal to the RHS. So let's simplify both sides:
LHS = sin^2x / (sinx + sinx*cosx)
RHS = (sin^2x - sinx*cosx) * cosx
To prove their equality, we can demonstrate that they are equivalent by showing that their numerators and denominators are equivalent:
Numerator equivalence:
sin^2x = sin^2x - sinx*cosx
Denominator equivalence:
sinx + sinx*cosx = cosx
We can show the numerator equivalence by adding sinx*cosx to both sides of the equation:
sin^2x + sinx*cosx = sin^2x - sinx*cosx + sinx*cosx
sin^2x + sinx*cosx = sin^2x
Next, we simplify the denominator equivalence by using the distributive property:
sinx + sinx*cosx = cosx*(1 + sinx) = cosx
Since both the numerator and denominator are equivalent on both sides, the LHS is indeed equal to the RHS:
LHS = sin^2x / (sinx + sinx*cosx) = (sin^2x - sinx*cosx) * cosx = RHS
Therefore, we have proven that (tanx)(sinx) / (tanx) + (sinx) = (tanx) - (sinx) / (tanx)(sinx).