Help with proving identities please

[nevag07]

i need help with proving these identities
1/sinxcosx - cosx/sinx= tanx

1/1+cosx=csc^2x-cscxcotx

sinx/1-cosx + 1_cosx/sinx = 2cscx

i would really appreciate this

 

[courtrigrad]

1. \frac{1}{\sin x \cos x} - \frac{\cos x}{\sin x} = \frac{\sin x - \sin x \cos^{2} x}{\sin^{2} x \cos x} = \frac{\sin x(1-\cos^{2}x)}{\sin x(\sin x \cos x)} = \frac{\sin x}{\cos x} = \tan x2. \frac{1}{1+\cos x} = \frac{1}{\sin^{2} x} - \frac{\cos x}{\sin^{2} x} = \frac{\sin^{2} x(1-\cos x)}{\sin^{2}x(\sin^{2}x)} = \frac{1}{1+\cos x}you do the last one.

you should come up with \frac{2-2\cos x}{\sin x(1-\cos x)} Now factor this, simplify, and see what you get.

 

[interested_learner]

Proof

Remember your triangles.

sine is y/hypotenuse

cosine is x/hypotenuse

tangent is y/x

and X^2 + y^2 = hypotenuse^2

Translate your problems into triangle form and the answers will pop out.

The problem with Courtrigrad's approach (while it is valid), is that he assumes the validity of the identity sin^2 X + cos^2 x = 1. When proving trig identities, it is usually more educational to go back to the basic definitions and not assume anything.

 

[d_leet]

Remember your triangles.


The problem with Courtrigrad's approach (while it is valid), is that he assumes the validity of the identity sin^2 X + cos^2 x = 1. When proving trig identities, it is usually more educational to go back to the basic definitions and not assume anything.

There is nothing wrong with Courtrigrad's method, I don't understand your claim that he assumes anything, I would take
sin²x + cos²x = 1 as a given because it is always true, and does follow from the basic definitions of the trigonometric functions. This is not the case with your method, however, your method to be technical will only work for angles strictly between 0 and 90 degrees and I wouldn't really call your definitions the basic ones.

 

[turdferguson]

As long as you don't carry things over and multiplying both sides like you would in algebra (you can find common denominators), that method is fine. The thing to remember is that you need to solve one side until you get to the other.