Question

1-sinxcosx/cosx(secx-cosecx)×sin2x-cos2x/sin3x+cos3x=sinx. prove it​

Answer 1

To prove:

$\dfrac{(1-sinxcosx)}{cosx(secx-cosecx)}\times \dfrac{sin^2x-cos^2x}{sin^3x+cos^3x}=sinx$

Proof:

First of all, have a look at the LHS(Left Hand Side) and try to solve it:$\dfrac{(1-sinxcosx)}{cosx(secx-cosecx)}\times \dfrac{sin^2x-cos^2x}{sin^3x+cos^3x}$

Let us use the algebraic formula:

1. $a^{2} -b^{2} =(a+b)(a-b)$

2. $a^{3} +b^{3} =(a+b)(a^2+b^2-ab)$

Here, $a = sinx$ and $b = cos x$, so the term from LHS:

$\dfrac{sin^2x-cos^2x}{sin^3x+cos^3x}\\\Rightarrow \dfrac{(sinx+cosx)(sinx-cosx)}{(sinx+cosx)(sin^2x+cos^2x-sinxcosx)}\\\Rightarrow \dfrac{(sinx-cosx)}{(sin^2x+cos^2x-sinxcosx)}\\\text{Using } sin^2\theta+cos^2\theta = 1:\\\\\Rightarrow \dfrac{(sinx-cosx)}{(1-sinxcosx)}$

Putting the value back to LHS:

$\dfrac{(1-sinxcosx)}{cosx(secx-cosecx)}\times \dfrac{sin^2x-cos^2x}{sin^3x+cos^3x}\\\Rightarrow \dfrac{(1-sinxcosx)}{cosx(secx-cosecx)}\times \dfrac{(sinx-cosx)}{(1-sinxcosx)}\\\Rightarrow \dfrac{(sinx-cosx)}{cosx(secx-cosecx)}$

Using the following identities:

$3.\ sec\theta = \dfrac{1}{cos\theta}\\4.\ cosec\theta = \dfrac{1}{sin\theta}$

$\Rightarrow \dfrac{(sinx-cosx)}{cosx(\dfrac{1}{cosx}-\dfrac{1}{sinx})}\\\Rightarrow \dfrac{(sinx-cosx)}{(\dfrac{cosx}{cosx}-\dfrac{cosx}{sinx})}\\\Rightarrow \dfrac{(sinx-cosx)}{(1-\dfrac{cosx}{sinx})}\\\Rightarrow \dfrac{sinx(sinx-cosx)}{(sinx-cosx)}\\\Rightarrow sinx = RHS$

Left Hand Side = Right Hand Side

Hence, proved that:

$\dfrac{(1-sinxcosx)}{cosx(secx-cosecx)}\times \dfrac{sin^2x-cos^2x}{sin^3x+cos^3x}=sinx$