Question

Simplify the first trigonometric expression by writing the simplified form in terms of the second expression (1/1-cosx)-(cos/1+cos);cscx

Answer 1

$(\frac{1}{1-cos(x)})-(\frac{cos(x)}{1+cos(x)})$

cscx is  1/sinx

maybe they want us to use pythagorean identity

$cos^2(x)+sin^2(x)=1$

I notice we have 1-cos(x) and 1+cos(x), if we multiply them, we get $1-cos^2(x)$

and if we look at the pythagorean identity and minus cos^2(x) from both sides, we get

$sin^2(x)=1-cos^2(x)$

since $csc(x)=\frac{1}{sin(x)}$, $csc^2(x)=\frac{1}{sin^2(x)}=\frac{1}{1-cos^2(x)}$

(recall that (a-b)(a+b)=a²-b²)

match the denomenators of the original fraction

multiply first fraction by $\frac{1+cos(x)}{1+cos(x)}$ and the 2nd by $\frac{1-cos(x)}{1-cos(x)}$

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

$((csc^2(x))(1+cos(x)))-((csc^2(x))(cos(x)-cos^2(x)))=$

$csc^2(x)+csc^2(x)cos(x)-(csc^2(x)cos(x)-csc^2(x)cos^2(x)=$

$csc^2(x)+csc^2(x)cos(x)-csc^2(x)cos(x)+csc^2(x)cos^2(x)=$

$csc^2(x)+csc^2(x)cos^2(x)=$

$csc^2(x)(1+cos^2(x))$, hmm, to get ride of those cos(x)

look to the pythagorean identity again

$sin^2(x)+cos^2(x)=1$, force one side into form 1+cos^2(x)

$1+cos^2(x)=2-sin^2(x)$, recall that since csc(x)=1/sin(x), sin(x)=1/csc(x) and sin^2(x)=1/(csc^2(x))

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

subsituting

$csc^2(x)(1+cos^2(x))=$

$(csc^2(x))(2-\frac{1}{csc^2(x)})=$ distributing

$2csc^2(x)-\frac{csc^2(x)}{csc^2(x)}=$

$2csc^2(x)-1$ is the simplified expression

Answer 2

Trigonometry identities are proved by using several other identities.

The simplified form of $\mathbf{\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}}$ is  $\mathbf{2csx^2 x - 1 }}$

The trigonometric expression is given as;

$\mathbf{\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}}$

Take LCM

$\mathbf{\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x} =\frac{1 + \cos x - \cos x(1 - \cos x)}{(1 - \cos x)(1 + \cos x)} }$

Open brackets

$\mathbf{\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x} =\frac{1 + \cos x - \cos x + \cos^2 x}{1 - \cos^2 x} }$

Evaluate like terms

$\mathbf{\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x} =\frac{1 + \cos^2 x}{1 - \cos^2 x} }$

In trigonometry:

$\mathbf{ \cos^2x = 1 - \sin^2 x}$

So, we have:

$\mathbf{\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x} =\frac{1 +1 - \sin^2 x}{sin^2 x} }}$

$\mathbf{\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x} =\frac{2 - \sin^2 x}{sin^2 x} }}$

Split

$\mathbf{\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x} =\frac{2}{sin^2x} - \frac{\sin^2 x}{sin^2 x} }}$

$\mathbf{\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x} =\frac{2}{sin^2x} - 1 }}$

The inverse of sin(x) is csc(x).

So, we have:

$\mathbf{\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x} =2csx^2 x - 1 }}$

Hence, the simplified form of $\mathbf{\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}}$ is  $\mathbf{2csx^2 x - 1 }}$

Read more about trigonometry identities at:

https://brainly.com/question/10680548