Proving Trigonometric Identity

[Kondwani Hauya]

May you check my work please? If you see anything wrong please help
Cos X cot x - 1 = csc x
1 - sin x
Solution
L.H.S = (1 - sin x) cot x - 1
1 - sin x
= cot r - 1= R.H.S

 

[Dr.Peterson]

I think you are saying that [MATH]\frac{\cos x\cot x}{1 - \sin x}-1 = \frac{(1-\sin x)\cot x}{1-\sin x} - 1 = \cot x - 1 = \csc x[/MATH].

No, that is not correct; it is not true that [MATH]\cos x = 1 - \sin x[/MATH], or that [MATH]\cot x - 1 = \csc x[/MATH].

Rather, [MATH]\cos^2 x = 1 - \sin^2 x[/MATH], and [MATH]\cot^2 x + 1 = \csc^2 x[/MATH].

I might start either by using a common denominator to obtain a single fraction, or writing cot x in terms of sin x and cos x.

 

[pka]

May you check my work please? If you see anything wrong please help
Cos X cot x - 1 = csc x
1 - sin x

Here is a second approach.
\(\displaystyle \begin{align*}\frac{\cos(x)\cot(x)}{1-\sin(x)}-1&=\frac{\cos(x)\frac{\cos(x)}{\sin(x)}}{1-\sin(x)}-1 \\&=\frac{\cos^2(x)}{\sin(x)(1-\sin(x))}-1\\&= \frac{1-\sin^2(x)}{\sin(x)(1-\sin(x))}-1\\&= \frac{1+\sin(x)}{\sin(x)}-1\end{align*}\)

Can you finish?

 

[Kondwani Hauya]

From your last step I finish like this by
using common denominator
1+ sin x - 1
sin x 1
1 + sin x - sin x
sin x
1/sin x = csc x

 

[pka]

From your last step I finish like this by
using common denominator
1+ sin x - 1
sin x 1
1 + sin x - sin x
sin x
1/sin x = csc x

Indeed, done.