1. ## identity

1. prove the following identity: $\displaystyle \frac{1+cosx}{sinx}+\frac{sinx}{1+cosx}=\frac{2}{s inx}$

2. prove the following identity: $\displaystyle sec^2x-2sec x cos x+ cos^2x=tan^2x-sin^2 x$

2. Hello william
Originally Posted by william
1. prove the following identity: $\displaystyle \frac{1+cosx}{sinx}+\frac{sinx}{1+cosx}=\frac{2}{s inx}$
$\displaystyle \frac{1+\cos x}{\sin x}+\frac{\sin x}{1+\cos x} = \frac{(1+\cos x)^2+\sin^2x}{\sin x (1+\cos x)}$

$\displaystyle =\frac{1+2\cos x +\cos^2x+\sin^2x}{\sin x(1+\cos x)}$

$\displaystyle =\frac{1+2\cos x +1}{\sin x(1+\cos x)}$

$\displaystyle =\frac{2(1+\cos x)}{\sin x(1+\cos x)}$

$\displaystyle =\frac{2}{\sin x}$

2. prove the following identity: $\displaystyle sec^2x-2sec x cos x+ cos^2x=tan^2x-sin^2 x$
$\displaystyle \sec^2x-2\sec x \cos x+ \cos^2x=(\sec x -\cos x)^2$

$\displaystyle = \left(\frac{1}{\cos x}-\cos x\right)^2$

$\displaystyle = \left(\frac{1-\cos^2x}{\cos x}\right)^2$

$\displaystyle = \frac{(1-\cos^2x)(1-\cos^2x)}{\cos^2 x}$

$\displaystyle = \frac{(\sin^2x)(1-\cos^2x)}{\cos^2 x}$

$\displaystyle = \frac{(\sin^2x-\sin^2x\cos^2x)}{\cos^2 x}$

$\displaystyle =\frac{\sin^2x}{\cos^2x}-\sin^2x$

$\displaystyle =\tan^2x -\sin^2x$