Question

Prove
cot2x.(secx-1/1+sinx) +sec2x(sinx-1/1+secx)=0

Answer 1

$\text{Consider,}$

$\displaystyle\,cot^2x(\frac{secx-1}{1+sinx})+sec^2x(\frac{sinx-1}{1+secx})$

$=\displaystyle\,\frac{cot^2x(secx-1)(secx+1)+sec^2x(sinx-1)(sinx+1)}{(1+secx)(1+secx)}$

$\text{Using,}\;\bf(a+b)(a-b)=a^2-b^2$

$=\displaystyle\,\frac{cot^2x(sec^2x-1)+sec^2x(sin^2x-1)}{(1+secx)(1+secx)}$

$=\displaystyle\,\frac{cot^2x(tan^2x)+sec^2x(-cos^2x)}{(1+secx)(1+secx)}$

$=\displaystyle\,\frac{cot^2x(tan^2x)-sec^2x(cos^2x)}{(1+secx)(1+secx)}$

$=\displaystyle\,\frac{\frac{1}{tan^2x}(tan^2x)-\frac{1}{cos^2x}(cos^2x)}{(1+secx)(1+secx)}$

$=\displaystyle\,\frac{1-1}{(1+secx)(1+secx)}$

$=\displaystyle\,\frac{0}{(1+secx)(1+secx)}$

$=0$

$\therefore\bf\,\displaystyle\,cot^2x(\frac{secx-1}{1+sinx})+sec^2x(\frac{sinx-1}{1+secx})=0$

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