Question

prove that sinx-cosx+1/sinx+cosx-1=1/secx-tanx

Answer 1

GIVEN :

Prove that $\frac{sinx-cosx+1}{sinx+cosx-1}=\frac{1}{secx-tanx}$

TO PROVE :

The following equation $\frac{sinx-cosx+1}{sinx+cosx-1}=\frac{1}{secx-tanx}$

SOLUTION :

Given that $\frac{sinx-cosx+1}{sinx+cosx-1}=\frac{1}{secx-tanx}$

We have to prove that the equation is true.

Now taking the LHS $\frac{sinx-cosx+1}{sinx+cosx-1}$

Let S=sinx and C=cosx

$\frac{S-C+1}{S+C-1}$

$=\frac{(S-C)+1}{(S+C)-1}$

Multiply and divide by (S-C)-1 we get,

$=\frac{(S-C)+1}{(S+C)-1}\times \frac{(S-C)-1}{(S-C)-1}$

By using the Algebraic identity :

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

$=\frac{(S-C)^2-1^2}{((S+C)-1)((S-C)-1)}$

$=\frac{S^2-2SC+C^2-1}{(S+C)(S-C)-S-C-S+C+1}$

By using the trignometric identity :

$sin^2x+cos^2x=1$ where sinx=C and cosx=C

$=\frac{S^2-2SC+C^2-(S^2+C^2)}{S^2-SC+CS-C^2-2S+1}$

$=\frac{S^2-2SC+C^2-S^2-C^2}{S^2-C^2-2S+(S^2+C^2)}$

$=\frac{-2SC}{2S^2-2S}$

$=\frac{-2SC}{2S(S-1)}$

$=\frac{-C}{S-1}$

$=\frac{-C}{-(1-S)}$

$=\frac{C}{1-S}$

$=\frac{cosx}{1-sinx}$

∴ $\frac{sinx-cosx+1}{sinx+cosx-1}=\frac{cosx}{1-sinx}\hfill (1)$

Now taking the RHS $\frac{1}{secx-tanx}$

$\frac{1}{secx-tanx}$

By using the trignometric identities

i) $secx=\frac{1}{cosx}$ and

ii) $tanx=\frac{sinx}{cosx}$

$=\frac{1}{\frac{1}{cosx}-\frac{sinx}{cosx}}$

$=\frac{1}{\frac{1-sinx}{cosx}}$

$=\frac{cosx}{1-sinx}$

∴ $\frac{1}{secx-tanx}=\frac{cosx}{1-sinx}\hfill (2)$

Comparing the equations (1) and (2) we get,

⇒ (1) = (2)

∴  LHS=RHS

∴  $\frac{sinx-cosx+1}{sinx+cosx-1}=\frac{1}{secx-tanx}$

Hence proved.